This commit is contained in:
KYOSG 2021-05-08 09:27:24 +08:00
parent bfe7d4905a
commit aeb9c19ba0
191 changed files with 9805 additions and 0 deletions

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/*
Description
z现在落在一个n*m的地图里,, z的方向感堪忧, (x,y),. 8: ,,,,,,,,. z的好朋友小y特别喜欢偶数, z能带回尽可能大的偶数之和.z请求你写个程序帮助他完成这个任务.
Input
n,m(1 <= n,m <= 1000) , n行,m个数字a[i][j],1 <= a[i][j] <= 10^9.x,y (1 <= x <= n, 1 <= y <= m)z现在所在的位置.
Output
,,z最多能带回的最大偶数之和
Sample
Input
3 4
1 2 3 4
1 2 3 4
1 2 3 4
2 2
Output
6
Hint
*/
#include<stdio.h>
int main()
{
long long int i,n,m,j,x,t=0,y,max=0;
long long map[1000][1000],sum[8]={0};
scanf("%lld %lld",&n,&m);
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
scanf("%lld",&map[i][j]);
}
}
scanf("%lld %lld",&x,&y);
x=x-1;
y=y-1;
if (map[y][x]%2==0)
{
t=map[y][x];
map[y][x]=1;
}
for(i=y;i<n;i++)//向下
{
if (map[i][x]%2==0)
{
sum[0]+=map[i][x];
map[i][x]=1;
}
}
for(i=y;i>=0;i--)//向上
{
if (map[i][x]%2==0)
{
sum[1]+=map[i][x];
map[i][x]=1;
}
}
for(i=x;i>=0;i--)//向左
{
if (map[y][i]%2==0)
{
sum[2]+=map[y][i];
map[y][i]=1;
}
}
for(i=x;i<m;i++)//向右
{
if (map[y][i]%2==0)
{
sum[3]+=map[y][i];
map[y][i]=1;
}
}
for(i=y,j=x;i>=0&&j<m;i--,j++)//向右上
{
if (map[i][j]%2==0)
{
sum[4]+=map[i][j];
map[i][j]=1;
}
}
for(i=y,j=x;i>=0&&j>=0;i--,j--)//向左上
{
if (map[i][j]%2==0)
{
sum[5]+=map[i][j];
map[i][j]=1;
}
}
for(i=y,j=x;i<n&&j>=0;i++,j--)//向左下
{
if (map[i][j]%2==0)
{
sum[6]+=map[i][j];
map[i][j]=1;
}
}
for(i=y,j=x;i<n&&j<m;i++,j++)//向右下
{
if (map[i][x]%2==0)
{
sum[7]+=map[i][x];
map[i][x]=1;
}
}
for(i=0;i<8;i++)
{
if(sum[i]>max)
{
max=sum[i];
}
}
if (t!=0)
{
max=max+t;
}
printf("%lld",max);
return 0;
}

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103
Test/2019ACM/守卫的DNA.c Executable file
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/*
Description
Ljn计算出了湖深度后Ljj为了奖励Ljn抓了稷下湖所有的鹅给Ljn吃py交易DNA序列DNA序列的字符串?DNA由AGCT?DNAimpossible
Input
n表示字符串的长度1n1000
n的字符串
Output
impossible
Sample
Input
8
AG?C??CT
Output
AGACGTCT
*/
#include<stdio.h>
#include<string.h>
int main()
{
char dna[1000],fl;
int a=0,c=0,g=0,t=0;
int len,i,max;
scanf("%d",&len);
getchar();
if(len%4!=0)
{
printf("impossible\n");
return 0;
}
max=len/4;
gets(dna);
for(i=0;i<len;i++)//计数
{
fl=dna[i];
switch(fl)
{
case 'A':
a++;
break;
case 'G':
g++;
break;
case 'C':
c++;
break;
case 'T':
t++;
break;
}
}
if (a>max||c>max||t>max||g>max)
{
printf("impossible\n");
return 0;
}
for(i=0;i<len;i++)
{
if (dna[i]=='?')
{
if (a<max)
{
dna[i]='A';
a++;
continue;
}
if (c<max)
{
dna[i]='C';
c++;
continue;
}
if (g<max)
{
dna[i]='G';
g++;
continue;
}
if (t<max)
{
dna[i]='T';
t++;
continue;
}
}
}
puts(dna);
printf("\n");
return 0;
}

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Test/2019ACM/想吃鹅肉? .c Executable file
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/*
Description
Ljn和船长Ljj乘坐塞尔号在稷下湖中滑水Ljn注意到了湖面上有一朵自闭的荷花H厘米Ljn又抓住荷花向正前航行了L厘米A点A点正上方Ljj要求Ljn计算湖的深度Ljn吃烤鹅肉Ljn为了能吃更多的鹅肉
Input
H和L(1H<L1000000)
Output
A点到湖面的距离
Sample
Input
1 2
Output
1.50
*/
#include<stdio.h>
int main()
{
long long int h,l;
double d;
scanf("%lld %lld",&h,&l);
d=(double)(l*l-h*h)/(2.0*h);
printf("%.2lf\n",d);
return 0;
}

139
Test/2019ACM/机器数.c Executable file
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/*
Description
180111
10
11001
+1
8
Input
T(1=<T<=255),T组数据
T行i1=<i<=T8
Output
Sample
Input
2
10101010
01010101
Output
11010110
01010101
Hint
00000000 10000000
*/
/*#include<stdio.h>
int main()
{
//此处不建议用int因为会有0开头的数会导致结果少一位本程序在第20行修补了这一错误
//应用字符串型并且因为在ASCII码表中数字是顺序的所以可直接运算
int i,num,t,T;
int n[8];
scanf("%d",&T);
while(T--)
{
scanf("%d",&num);
t=num;
for(i=7;t>0;i--)
{
n[i]=t%10;
t=t/10;
}
if(i==0)//修补因为0开头的数字导致位数减少的问题
{
n[0]=0;
}
if (n[0]==1)
{
for(i=1;i<8;i++)
{
n[i]=!n[i];//取反
}
for(i=7;i>0;i--)
{
if(i==7)
{
n[i]=n[i]+1;
}
if (n[i]==2)//逢2进1
{
n[i]=0;
n[i-1]=n[i-1]+1;//进位
}
}
}
for(i=0;i<8;i++)
{
printf("%d",n[i]);
}
printf("\n");
}
return 0;
}
*/
#include <stdio.h>
int main()
{
char a[8];
int t,i;
scanf("%d",&t);
while (t--)
{
scanf("%s",a);
if (a[0] == '1')
{
for (i = 1; i < 8; i++)
{
if (a[i] == '1')
{
a[i] = '0';
}
else
{
a[i] = '1';
}
}
}
if (a[0] == '1')
{
for (i = 7; i >= 0 ; i--)
{
if (i == 7)
{
a[i] = a[i] + 1;
}
if (a[i] == '2')
{
a[i] = '0';
a[i - 1] = a[i - 1] + 1;
}
}
}
for (i = 0;i < 8;i++)
{
printf("%c",a[i]);
}
printf("\n");
}
return 0;
}

76
Test/2019ACM/生化危机.c Executable file
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/*
Description
BHS公司制造了一种病毒ljj当然会阻止这种行为 W (),B() ljj
Input
NN行 (1<= N <=115)
N行字符串M1 <= M <= 115
N行M列中包含两个字母(W,B)
Output
x,y(1<=x<=N,1<=y<=M)
Sample
Input
5
WWBBBW
WWBBBW
WWBBBW
WWWWWW
WWWWWW
Output
2 4
*/
#include<stdio.h>
#include<string.h>
int main()
{
int n,i,j,up=0,left=0,down=0,right=0;
char str[115][115];
int x,y;
long int len;
scanf("%d",&n);
getchar();//吃回车
for(i=0;i<n;i++)
{
gets(str[i]);//自动换列
//scanf("%s",str[i]);也可
}
len = strlen(str[0]);
for(i=0;i<n;i++)
{
for(j=0;j<len;j++)
{
if (str[i][j]=='B')
{
if (!up && !left)//! 代表值得取反对于整形变量只要不为0使用 ! 取反都是00取反就是1
{
left = j+1;
up = i+1;
}
right = j + 1;
down = i+1;
}
//循环结束后便记录下了最后一个点的位置
}
}
x=(right+left)/2;
y=(down+up)/2;
printf("%d %d\n",y,x);
return 0;
}

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/*
Description
12436546 2 + 4 + 6 + 4 + 6
Input
n (0 <= n <= 2147483647)
Output
n
Sample
Input
6768
Output
20
*/
#include<stdio.h>
int main()
{
int n, temp, i, temp2, c=0;
scanf("%d", &n);
i = 0;
while (n>=10)
{
temp=0;
temp2=0;
temp = n%10;
temp2 = temp%2;
if (temp2 == 0)
{
i = i+temp;
}
else
c = c+1;
n = n/10;
}
c = 0;
c = n%2;
if (c == 0)
{
i = i + n;
printf("%d\n",i);
}
else
printf("%d\n", i);
return 0;
}

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/*
Description
24 "HH:mm" 05:20 12 "h:mm AM/PM" "5:20 AM"
24 12
0 12 (AM)
1~11 1~11 (AM)
12 12 (PM)
13~23 1~11 (PM)
"00:00" "12:00 AM""01:20" "1:20 AM""12:35" "12:35 PM""13:17" "1:17 PM""23:59" "11:59 PM"
24 12
Input
24
Output
12 0
Sample
Input
00:05
Output
12:05 AM
Hint
使 scanf("%d:%d") 使 printf("%d:%02d")
*/
#include<stdio.h>
int main ()
{
int h1, m1;
int h2;
scanf("%d:%02d", &h1, &m1);
h2 = h1 - 12;
if (h1 == 12)
printf("%d:%02d PM\n", h1, m1);
else if (h1 == 0)
printf("%d:%02d AM\n", -h2, m1);
else
{
if (h2 < 0)
printf("%2d:%02d AM\n", h1, m1);
else
printf("%2d:%02d PM\n", h2, m1);
}
return 0;
}

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#include<stdio.h>
#include<string.h>
int main()
{
int n,i,j;
char s[3][110],t[110];
for(i=0;i<3;i++)
scanf("%s",s[i]);
for(i=0;i<2;i++)
for(j=i+1;j<3;j++)
if(strcmp(s[i],s[j])>0)
{
strcpy(t,s[i]);
strcpy(s[i],s[j]);
strcpy(s[j],t);
}
for(i=0;i<3;i++)
{
if(i==2) printf("%s\n",s[i]);
else printf("%s ",s[i]);
}
return 0;
}

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/*
Description
Input
Output
ASCII减去32ASCII加上32
A => a -32
a => A +32
*/
#include <stdio.h>
int main()
{
char a;
a = getchar();
a = a - 32;
putchar(a);
return 0;
}

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/*
Description
10 3
10 3
Input
double范围内的正实数 a
Output
Sample
Input
123.56789
Output
123.567890
* 123.568*
*123.568 *
*/
#include <stdio.h>
int main()
{
double a;
scanf("%lf",&a);
printf("%lf\n",a);
printf("*%10.3lf*\n",a);
printf("*%-10.3lf*\n",a);
return 0;
}

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/*
Description
long long
Input
long long
Output
long long
*/
#include <stdio.h>
int main()
{
long long a,b,c;
scanf("%lld %lld", &a, &b);
c = a + b;
printf("%lld",c);
return 0;
}

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/*
Description
'
" 包裹。
Input
Output
Sample
Input
A
Output
'A'
"A"
*/
#include <stdio.h>
int main ()
{
char a;
scanf("%c",&a);
printf("\'%c\'\n",a);
printf("\"%c\"",a);
return 0;
}

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/*
Description
`` \ `` a b c
Input
int范围内的正整数 `` \ ``
Output
Sample
Input
9\17\2018
Output
9\17\2018
*/
#include<stdio.h>
int main()
{
int a, b, c;
char d, e;
scanf("%d %c %d %c %d",&a, &d, &b, &e, &c);
printf("%d%c%d%c%d", a, d, b, e, c);
return 0;
}

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/*
Description
`` \ `` a b c
Input
int范围内的正整数 `` \ ``
Output
Sample
Input
9\17\2018
Output
9\17\2018
*/
#include<stdio.h>
int main()
{
int a, b, c;
char d, e;
scanf("%d %c %d %c %d",&a, &d, &b, &e, &c);
printf("%d%c%d%c%d", a, d, b, e, c);
return 0;
}

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/*
Description
Input
int范围内的正整数 a
Output
Sample
Input
456
Output
456
1c8
1C8
*/
#include<stdio.h>
int main()
{
int a;
scanf("%d",&a);
printf("%d\n",a);
printf("%x\n",a);
printf("%X\n",a);
return 0;
}

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/*
Description
8 0
8
Input
int范围内的正整数 a
Output
Sample
Input
123
Output
123
*00000173*
*173 *
*/
#include<stdio.h>
int main()
{
int a;
scanf("%d",&a);
printf("%d\n",a);
printf("*%08o*\n",a);
printf("*%-8o*\n",a);
return 0;
}

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/*
Description
8 0
8
Input
int范围内的正整数 a
Output
Sample
Input
123456
Output
123456
*00123456*
*123456 *
*/
#include<stdio.h>
int main()
{
int a;
scanf("%d",&a);
printf("%d\n",a);
printf("\*%08d\*\n",a);
printf("\*%-8d\*\n",a);
}

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/*
Description
8
8
Input
int范围内的正整数 a
Output
Sample
Input
123456
Output
123456
* 123456*
*123456 *
*/
#include<stdio.h>
int main()
{
int a;
scanf("%d", &a);
printf("%d\n", a);
printf("*%8d*\n", a);
printf("*%-8d*\n", a);
return 0;
}

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#include<stdio.h>
int main()
{
printf("Hello World!");
return 0;
}

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/*
Description
4 3
4 3
Input
Output
Sample
Input
c
Output
c
* c*
*c *
*/
#include<stdio.h>
int main()
{
char a;
scanf("%c", &a);
printf("%c\n", a);
printf("*%4c*\n", a);
printf("*%-4c*\n", a);
return 0;
}

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#include<stdio.h>
int main()
{
int a;
int b;
int c;
scanf("%d %d", &a, &b);
c = a + b;
printf("%d",c);
return 0;
}

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#include<stdio.h>
int main()
{
int x,y, z;
scanf("%d %d", &x, &y);
z = x;
x = y;
y = z;
printf("%d %d",x,y);
return 0;
}

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/*
Description
Input
3
Output
*/
#include <stdio.h>
int main()
{
int a, b, c, d, e;
scanf("%d", &a);
b = a / 100;
c = (a / 10 - b * 10);
d = a - b * 100 - c * 10;
e = (d * 100) + (c * 10) + b;
printf("%d", e);
return 0;
}

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/*
Description
3n元钱
Input
n元n为整数
Output
*/
#include <stdio.h>
int main()
{
int m,n,t,c;
scanf("%d", &m);
t = m*10;
c = t % 3;
n = t / 3;
printf("%d %d", n, c);
return 0;
}

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#include<stdio.h>
int main()
{
int a,b,c,d,e;
float f;
scanf("%d %d %d", &a, &b, &c);
d = a + b + c;
e = a * b * c;
f = (d + 0.00) / 3;
printf("%d %d %.2f", d, e, f);
return 0;
}

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#include<stdio.h>
int main ()
{
float pi = 3.1415926;
int r,h;
float L,S,SC,V;
scanf("%d %d", &r, &h);
L = pi * r * 2;
S = pi * r * r;
SC = L*h;
V = S * h;
printf("%.2f %.2f %.2f %.2f", L, S, SC, V);
return 0 ;
}

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#include<stdio.h>
int main()
{
float C;
float F;
scanf("%f",&F);
C = 5*( F - 32 ) / 9;
printf("%.2f",C);
return 0;
}

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/*
Description
getchar()putchar
Input
Output
*/
#include<stdio.h>
int main ()
{
int a;
a = getchar();
putchar(a);
return 0;
}

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/*
Description
Input
3
Output
Sample
Input
1 2 3
Output
2
*/
#include <stdio.h>
int main ()
{
int a, b, c, t;
scanf("%d %d %d", &a, &b, &c);
if ( a < b )
{
//交换ab的数值
t = a;
a = b;
b = t;
}
if ( b < c )
{
//交换bc的数值
t = b;
b = c;
c = t;
}
if (b > a)
{
//交换ac的数值
t = b;
b = a;
a = t;
}
printf ("%d\n", b);
return 0;
}

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/*
Description
n能否同时被3和5整除
Input
n
Output
35YesNo
Sample
Input
15
Output
Yes
*/
#include <stdio.h>
int main ()
{
int a;
scanf("%d", &a);
if (( a % 3 == 0) && (a % 5 == 0) )
printf("Yes");
else
printf("No");
return 0;
}

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/*
Description
Input
year
Output
YesNo
Sample
Input
2000
Output
Yes
*/
#include <stdio.h>
int main ()
{
int year;
scanf("%d", &year);
if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
printf("Yes\n");
else
printf("No\n");
return 0;
}

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/*
Description
Input
+-*/)
0
Output
Sample
Input
30 50
*
Output
1500
*/
#include <stdio.h>
int main ()
{
int a, b, sum;
char c;
scanf("%d %d", &a, &b);
scanf(" %c", &c);
if (a != 0 && b != 0)
{
if ('+' == c)
{
sum = a + b;
printf("%d", sum);
}
else if ('-' == c)
{
sum = a - b;
printf("%d", sum);
}
else if ('*' == c)
{
sum = a * b;
printf("%d", sum);
}
else if ('/' == c)
{
sum = a / b;
printf("%d", sum);
}
}
else return 0;
return 0;
}

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/*
Description
a,b,c
Input
Output
Sample
Input
1 2 3
Output
5
*/
#include<stdio.h>
#include<stdlib.h>
int main()
{
int y,m,d;
scanf("%d\\%d",&y,&m);//输入\时用\\转义
switch(m)
{
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:d=31;break;
case 4:
case 6:
case 9:
case 11:d=30;break;
case 2:
if((y%4==0&&y%100!=0)||(y%400==0))
d=29;
else
d=28;
}
printf("%d\n",d);
return 0;
}

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/*
Description
1 Monday
2 Tuesday
3 Wednesday
4 Thursday
5 Friday
6 Saturday
7 Sunday
Input
1-7
Output
Sample
Input
2
Output
Tuesday
*/
#include <stdio.h>
int main ()
{
int nu;
scanf("%d", &nu);
switch (nu)
{
case 1:
printf("Monday");
break;
case 2:
printf("Tuesday");
break;
case 3:
printf("Wednesday");
break;
case 4:
printf("Thursday");
break;
case 5:
printf("Friday");
break;
case 6:
printf("Saturday");
break;
case 7:
printf("Sunday");
break;
default:
break;
}
return 0;
}

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/*
Description
(24::
3525--030525.
Input
1
2
Output
::
Sample
Input
12:01:12
13:09:43
Output
01:08:31
*/
#include <stdio.h>
int main()
{
int a;
scanf("%d", &a);
if (a<0)
printf("%d", -a);
else
printf("%d", a);
return 0;
}

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/*
Description
(24::
3525--030525.
Input
1
2
Output
::
Sample
Input
12:01:12
13:09:43
Output
01:08:31
*/
#include<stdio.h>
int main()
{
int a,b,d,e,f,g,i, j, k, s2, m2, h2;
char c;
scanf("%d %c %d %c %d", &a, &c, &b, &c, &d);
scanf("%d %c %d %c %d", &e, &c, &f, &c, &g);
//大小置换
if (a < e)
{
h2 = a;
a = e;
e = h2;
m2 = b;
b = f;
f = m2;
s2 = d;
d = g;
g = s2;
}
else
{
a = a;
b = b;
d = d;
e = e;
f = f;
g = g;
}
i = a - e;
j = b - f;
k = d - g;
// 秒
if (k < 0)
{
k = 60 + k;
j = j - 1;
}
else
k = k;
// 分
if (j < 0)
{
j = 60 + j;
i = i - 1;
}
else if (j >= 60)
{
j = j - 60;
i = i + 1;
}
else if (j >=0 && j <= 60)
j = j;
// 时
if (i < 0)
i = - i;
else
i = i;
printf("%02d%c%02d%c%02d", i, c, j, c, k);
return 0;
}

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/*
Description
Input
5 9
Output
max=9
Sample
Input
5 9
Output
max=9
*/
#include<stdio.h>
int main ()
{
int a, b;
scanf("%d %d", &a, &b);
if (a>b)
printf("max=%d", a);
else
printf("max=%d", b);
return 0;
}

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/*
Description
xy
Input
xy(1 <= (x, y) <= 1000)
Output
Sample
Input
100 200
Output
200
*/
#include<stdio.h>
int main ()
{
int x, y;
scanf("%d %d", &x, &y);
if (x>y)
printf("%d", x);
else
printf("%d", y);
return 0;
}

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/*
Description
1000000
Input
abab>0a+b口算出的答案
Output
YESNO
Sample
Input
1 2
3
Output
YES
*/
#include<stdio.h>
int main ()
{
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
if (c == a + b)
printf("YES\n");
else
printf("NO\n");
return 0;
}

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/*
Description
0101I like youHe he
Input
01
Output
I like youHe he
Sample
Input
0
Output
He he
Hint
*/
#include<stdio.h>
int main()
{
int a;
scanf("%d", &a);
if (a == 1)
printf("I like you\n");
else if (a == 0)
printf("He he\n");
return 0;
}

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/*
Description
Input
Output
Sample
Input
5,7,9
Output
max=9
*/
#include<stdio.h>
int main ()
{
int a,b,c,t,max;
scanf("%d,%d,%d",&a,&b,&c);
if(a>b)
{
t=a;
a=b;
b=t;
}
if(a>c)
{
t=a;
a=c;
c=t;
}
if(b>c)
{
t=b;
b=c;
c=t;
}
max=c;
printf("max=%d\n",max);
return 0;
}

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/*
Description
a,b,c
Input
Output
Sample
Input
1 2 3
Output
5
*/
#include<stdio.h>
int main()
{
int a, b, c, d, e, f, t, max;
scanf("%d %d %d",&d,&e,&f);
a = d + e;
b = d + f;
c = e + f;
if(a>b)
{
t=a;
a=b;
b=t;
}
if(a>c)
{
t=a;
a=c;
c=t;
}
if(b>c)
{
t=b;
b=c;
c=t;
}
max=c;
printf("%d\n",max);
return 0;
return 0;
}

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/*
Description
abc
Input
abc
Output
abc的值
Sample
Input
4 3 5
Output
5 4 3
*/
#include <stdio.h>
int main ()
{
int a, b, c, t;
scanf("%d %d %d", &a, &b, &c);
if ( a < b )
{
//交换ab的数值
t = a;
a = b;
b = t;
}
if ( b < c )
{
//交换bc的数值
t = b;
b = c;
c = t;
}
if (b > a)
{
//交换ac的数值
t = b;
b = a;
a = t;
}
printf ("%d %d %d\n", a, b, c);
return 0;
}

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/*
Description
Your task is to Calculate a + b.
Too easy?! Of course! I specially designed the problem for acm beginners.
You must have found that some problems have the same titles with this one, yes, all these problems were designed for the same aim
Input
The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.
Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
Sample
Input
1 5
10 20
Output
6
30
*/
#include<stdio.h>
int main()
{
int a,b;
while(scanf("%d %d", &a,&b) != EOF)
{
printf("%d\n",a+b);
}
return 0;
}

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/*
Description
Your task is to Calculate a + b.
Input
Inputcontains multiple test cases. Each test case contains a pair of integers a and b, one pair of integers per line. A test case containing 0 0 terminates the input and this test case is not to be processed.
Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
Sample
Input
1 5
10 20
0 0
Output
6
30
*/
#include<stdio.h>
int main()
{
int a,b;
while(scanf("%d %d", &a,&b) !=EOF)
{
if (a!=0||b!=0)
printf("%d\n",a+b);
else
break;
}
return 0;
}

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/*
Description
Your task is to Calculate a + b.
Input
The input will consist of a series of pairs of integers a and b, separated by a space, one pair of integers per line.
Output
For each pair of input integers a and b you should output the sum of a and b, and followed by a blank line.
Sample
Input
1 5
10 20
Output
6
30
*/
#include<stdio.h>
int main()
{
int a,b;
while(scanf("%d %d", &a,&b) !=EOF)
{
printf("%d",a+b);
printf("\n");
printf("\n");
}
return 0;
}

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/*
Description
Input
Output
Sample
Input
1
1.5
Output
4.189
14.137
Hint
PI = 3.1415927
#define PI 3.1415927
*/
#include<stdio.h>
#define PI 3.1415927
int main()
{
double r;
double v=0;
while (scanf("%lf", &r)!=EOF)
{
v = (4.0*PI*r*r*r)/3.0;
printf("%.3lf\n", v);
}
return 0;
}

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/*
Description
Input
Output
Sample
Input
1
1.5
Output
4.189
14.137
Hint
#define PI 3.1415927
*/
#include <stdio.h>
#define pi 3.1415927
int main ()
{
double r1, v1;
while (scanf("%lf", &r1) != EOF)
{
v1 = (pi * r1 * r1 * r1) * (double)4/3;
printf("%.3lf\n", v1);
}
return 0;
}

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/*
Description
X是一个勤劳的小孩L的绳子能晾开多少件宽为W的衣服
Input
L,W
Output
Sample
Input
10 5
10 4
Output
2
2
*/
#include <stdio.h>
int main ()
{
int l, w, n;
while (scanf("%d%d", &l, &w) != EOF)
{
n = l/w;
printf("%d\n", n);
}
return 0;
}

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/*
Description
SuShan过年要给孩子们发压岁钱喽SuShan
SuShan从瑞士银行提出1000000来给孩子们分SuShan希望你能帮他计算一下每个孩子给多少钱
Input
T
T N1<= N <= 10000000
Output
No
Sample
Input
3
1
2
3
Output
1000000
500000
No
*/
#include <stdio.h>
int main()
{
int n, x;
double a;
int money = 10000000;
scanf("%d", &x);
while (x--)
{
scanf("%d", &n);
if (1 <= n && money >=n)
{
a = (double)money / n;
if (a != (int)a)
{
printf("No\n");
}
else
printf("%.0lf\n", a);
}
else
{
return 0;
}
}
return 0;
}
#include<stdio.h>
int main()
{
int T, N,c,a;
T= 0;
scanf("%d",&T);
while (T>0)
{
scanf("%d",&N);
a = 1000000%N;
c = 1000000/N;
if (a==0)
printf("%d\n",c);
else
printf("NO\n");
T=T-1;
}
return 0;
}
#include <stdio.h>
int main()
{
int money=1000000;
int i;
int n;
scanf("%d",&i);
while(i--)
{
scanf("%d",&n);
if (money % n == 0)
printf("%d\n",money/n);
else
printf("No\n");
}
return 0;
}

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/*
Description
1{1,2,3,4......}
n 1 - n
Input
n 1 <= n <= 1000
Output
Sample
Input
2
Output
3
*/
#include<stdio.h>
int main ()
{
int n, i;
int sum = 0;
scanf("%d", &n);
i = 1;
while (i<=n)
{
sum = sum + i;
++i;
}
printf("%d\n",sum);
return 0;
}

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/*
Description
NN^3
Input
N的值N<=1024
Output
Sample
Input
3
Output
351
*/
#include <stdio.h>
int main()
{
int n, s;
scanf("%d", &n);
s = n * n * n;
printf("%d", s);
return 0;
}

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/*
Description
NN^3
Input
N的值N<=1024
Output
Sample
Input
3
Output
351
*/
#include <stdio.h>
int main()
{
int n, c, a, sum=0;
scanf("%d", &n);
n = n * n * n;
while(n>0)
{
a = n % 10;
n = n / 10;
c = a * a * a;
sum = sum + c;
}
printf("%d\n",sum);
}

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/*
Description
n cmk cmm cm
Input
n, m, k ( 0 <= n<= 10000, 0 <= m <= 10000,0 <= k <= 10000)
Output
Sample
Input
100 200 5
Output
20
*/
#include <stdio.h>
int main()
{
int n,m,k,t=0;
scanf("%d %d %d",&n,&m,&k);
while(m>n)
{
m = m - k;
t++;
}
printf("%d\n",t);
return 0;
}

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/*
Description
12436546 2 + 4 + 6 + 4 + 6
Input
n (0 <= n <= 2147483647)
Output
n
Sample
Input
6768
Output
20
*/
#include <stdio.h>
int main ()
{
int n, i, a, j;
scanf("%d", &n);
i = 0;
while (n > 0)
{
a = n %10;
n = n / 10;
j = a %2;
if (j == 0)
i = i + a;
else
n = n;
}
printf("%d\n", i);
return 0;
}

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/*
Description
n m ( m [0,9]) m n
n = 2122345 , m = 2 3 2 2122345
Input
n m 0 <= m <= 91 <= n <= 2147483647
Output
m n
Sample
Input
2122345 2
Output
3
*/
#include <stdio.h>
int main()
{
int n, m, a, i;
scanf("%d %d", &n, &m);
if ((0 <= m && m <= 9)&&(1 <= n && n <= 214748364))
{
i = 0;
while (n > 0)
{
a = n%10;
n = n/10;
if (a == m)
{
++i;
}
else
n = n;
}
}
else
return 0;
printf("%d\n", i);
return 0;
}

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/*
Description
C语言的学习是我们程序设计基础的重点和主要内容
Input
nn >= 0n < 0退
Output
YesNo
Sample
Input
100
-100
Output
Yes
No
*/
#include<stdio.h>
int main()
{
int n;
while(scanf("%d", &n) != EOF)
{
if (n >= 0)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}

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/*
Description
3
Input
:
3
Output
YesNo
Sample
Input
2
1 2 3
1 4 4
Output
No
Yes
*/
#include <stdio.h>
int main ()
{
int a, n1, n2, n3;
int e, s;
scanf("%d", &a);
while (scanf("%d %d %d", &n1, &n2, &n3)!= EOF)
{
s = n1 + n2 + n3;
e = (float)s / 3;
if ((n1 > e && n2 > e) || (n1 > e && n3 >e)||(n2 > e && n3 >e))
printf("Yes\n");
else
printf("No\n");
}
return 0;
}

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/*
Description
F(x) = x^2 + 1 x> 0
F(x) = -x x<0
F(x) = 100.0 x=0
xx为实数
Input
x
Output
xF(x),1
Sample
Input
8.00
-5.0
Output
65.0
5.0
*/
#include <stdio.h>
int main ()
{
double x;
while (scanf("%lf", &x) != EOF)
{
if (0 < x )
{
x = x * x +1;
printf("%.1lf\n", x);
}
else if (0 > x)
{
x = -x;
printf("%.1lf\n", x);
}
else
{
x = 100;
printf("%.1lf\n", x);
}
}
return 0;
}

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/*
Description
Your task is to Calculate a + b.
Input
Your task is to Calculate a + b.
Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
Sample
Input
2
1 5
10 20
Output
6
30
*/
#include<stdio.h>
int main()
{
int a, b, sum, i;
for (scanf("%d", &i); scanf("%d %d", &a, &b)!=EOF; i--)
{
sum = a+b;
printf("%d\n", sum);
}
return 0;
}

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/*
Description
N个整数
Input
NN<=100
N个用空格隔开的整数Ti (1 <= i <= N, 0 <= Ti <= 10000000)
Output
Sample
Input
5
1 2 3 5 4
Output
5 1 3
*/
#include<stdio.h>
int main()
{
int n=0;
long long int max=0,min=10000000,sum=0,num=0,average;
int i=1;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%lld",&num);
if(num>max)
max=num;
if(min>num)
min=num;
sum=sum + num;
}
average=sum/n;
printf("%lld %lld %lld",max,min,average);
return 0;
}

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/*
Description
Input
int
Output
int
Sample
Input
64 48
Output
16
192
*/
#include<stdio.h>
int main()
{
int a,b,i,temp;
int max,min;
scanf("%d %d",&a, &b);
if (b>a)
{
temp=a;
a=b;
b=temp;
}
else
{
a=a;
b=b;
}
i=a;
while(i>0)//最大公因数
{
if ((a%i==0)&&(b%i==0))
{
// printf("%d\n",i);
min=i;
break;
}
else
i--;
}
//最小公倍数是两数乘积除以最大公约数
max=a*b/min;
printf("%d\n%d\n",min, max);
return 0;
}

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/*
Description
"This is a prime."
This is not a prime.
Input
n(1 <= n <= 1000000)
Output
n是否为素数
n是素数则输出"This is a prime."
This is not a prime.
11
Sample
Input
3
Output
This is a prime.
*/
#include <stdio.h>
int main()
{
int a,i,b,c=0;
scanf("%d",&a);
if (1==a)
{
printf("This is not a prime.\n");
}
else
{
for (i=2; i<a; i++)
{
b= a % i;
if (b==0)
{
c++;
}
else
{
c=c;
}
}
if (0==c)
{
printf("This is a prime.\n");
}
else
{
printf("This is not a prime.\n");
}
}
return 0;
}

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/*
Description
n个整数中的绝对值最大的数
Input
2nn个整数
Output
n个整数中绝对值最大的数
Sample
Input
5
-1 2 3 4 -5
Output
-5
*/
#include<stdio.h>
int main()
{
int n=0,t = 0;
int max = 0,num=0,temp = 0;
int i=1;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&num);
if (num<0)
{
temp=-num;
}
else
{
temp=num;
}
if(temp>max)
{
max=temp;
t=num;
}
}
printf("%d\n",t);
return 0;
}

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/*
Description
n值
Input
n值
Output
5
Sample
Input
1
Output
2.66667
*/
#include<stdio.h>
int main()
{
int n = 0,i;
double pi = 0.0;
scanf("%d",&n);
for (i=1; i<=n; i++)
{
pi=pi+(1.0/(4*i-3));
pi=pi-(1.0/(4*i-1));
}
printf("%.5lf\n",4*pi);
return 0;
}

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/*
Description
1*1=12*2=4 n n*n N
Input
N 0 <= N <= 10000
Output
Sample
Input
15
Output
3
*/
#include <stdio.h>
int main()
{
int
return 0;
}

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/*
Description
1*1=12*2=4 n n*n N
Input
N 0 <= N <= 10000
Output
Sample
Input
15
Output
3
*/
#include <stdio.h>
int main()
{
int n,i,sum,d=0;
scanf("%d",&n);
for (i=1;n>0; i++)
{
sum=i*i;
n=n-sum;
if (n<0)
{
break;
}
d=d+1;
}
printf("%d\n",d);
return 0;
}

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/*
Description
2/1, 3/2, 5/3, 8/5, 13/8, n项之和
Input
n1n10
Output
n项和6
Sample
Input
3
Output
5.166667
*/
#include<stdio.h>
int main()
{
int x=1,y=2,n = 0,i,t;
double sum=0.0;
scanf("%d", &n);
for (i=0; i<n; i++)
{
sum= sum+(double)y/x;
t = y;
y = x+y;
x=t;
}
printf("%.6lf\n",sum);
return 0;
}

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/*
Description
5A,B,C,D和E
9090A,80~9080B,70~8070C
60~7060DE
Input
NN<= 100N行数据每行一个整数0~100
Output
A nA
B nB
C nC
D nD
E nE
A,B,C,D,E代表等级nAnB等代表个等级的人数
Sample
Input
6
66
73
85
99
100
59
Output
A 2
B 1
C 1
D 1
E 1
*/
#include<stdio.h>
int main()
{
int i,na = 0,nb = 0,nc = 0,nd = 0 ,ne= 0 ,mark = 0;
for (scanf("%d",&i); i>0; i--)
{
scanf("%d",&mark);
if (mark>=90)
{
na++;
}
else if(mark>=80)
{
nb++;
}
else if(mark>=70)
{
nc++;
}
else if(mark>=60)
{
nd++;
}
else
{
ne++;
}
}
printf("A %d\nB %d\nC %d\nD %d\nE %d\n",na,nb,nc,nd,ne);
return 0;
}

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/*
Description
Your task is to Calculate a + b.
Input
Your task is to Calculate a + b.
Output
For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.
Sample
Input
2
1 5
10 20
Output
6
30
*/
#include <stdio.h>
int main()
{
int a, b, n, sum;
for (scanf("%d", &n); n > 0; n--)
{
scanf("%d %d", &a, &b);
sum = a + b;
printf("%d\n", sum);
}
return 0;
}

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/*
Description
n1n的和
Input
n
Output
1n的和
Sample
Input
3
Output
6
*/
#include <stdio.h>
int main ()
{
int n, sum;
scanf("%d", &n);
for (sum = 0; n > 0; n--)
{
sum = sum + n;
}
printf("%d\n", sum);
return 0;
}

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/*
Description
0nn的阶乘
0 1
Input
0n
Output
n
Sample
Input
3
Output
6
*/
#include <stdio.h>
int main ()
{
int n, count;
scanf("%d", &n);
if (n != 0 && n > 0)
{
for (count = 1; n > 0; n--)
{
count = count * n;
}
printf("%d\n", count);
}
else if (0 == n)
{
printf("1\n");
}
else
return 0;
return 0;
}

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/*
Description
2
Input
NN行分别是两个待比较的整数
Output
N行
Sample
Input
2
1 2
15 10
Output
2
15
*/
#include <stdio.h>
int main ()
{
int n, a, b;
for (scanf("%d", &n); 0<n; n--)
{
scanf("%d %d", &a, &b);
if (a>b)
{
printf("%d\n", a);
}
else
{
printf("%d\n", b);
}
}
return 0;
}

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/*
Description
C语言编写一个程序NN次N*(1->N)5
5*1=5
5*2=10
5*3=15
5*4=20
5*5=25
Input
NN<=100
Output
N行数据
Sample
Input
5
Output
5*1=5
5*2=10
5*3=15
5*4=20
5*5=25
*/
#include<stdio.h>
int main()
{
int sum=1,i, n;
scanf("%d",&n);
for (i=1; n>=i; i++)
{
sum = n*i;
printf("%d\*%d\=%d\n", n, i, sum);
}
return 0;
}
/*
#include <stdio.h>
int main()
{
int n;
int i;
int y;
scanf("%d",&n);
for(i = 1; i <= n; i++)
{
y = n * i;
printf("%d*%d=%d\n",n,i,y);
}
return 0;
}
*/

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/*
Description
s=a+aa+aaa+aaaa++aaaa(n位
a的值由键盘输入n也由键盘输入
Input
a的值
n
Output
n个数完成求和运算后的结果
a=3n=6s=3+33+333+3333+33333+333333
Sample
Input
3
6
Output
370368
*/
#include<stdio.h>
int main ()
{
int a, n, c;
int s = 0, i = 0;
scanf ("%d", &a);
scanf("%d", &n);
for (c=a; i<n; ++i)
{
s = s + c;
c = c * 10 + a;
}
printf("%d\n",s);
return 0;
}

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/*
Description
n m
491625n m 0 100000000 nm不一定有序
Input
T
T n,m (0 <= n, m <= 100000000)
Output
Sample
Input
3
1 4
10 3
17 20
Output
5
13
0
*/
#include<stdio.h>
#include<math.h>
int main()
{
int T,i,n,m,s,k,t;
scanf("%d",&T);
while(T--)
{
s=0;
scanf("%d %d",&n,&m);
if(m<n)
{
t=m;
m=n;
n=t;
}
for(i=n; i<=m; i++)
{
for(k=1; k<=sqrt(i); k++)
{
if(k*k==i)s=s+i;
}
}
printf("%d\n",s);
}
return 0;
}

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/*
Description
n(1n9)
Input
n1n9
Output
n-1n-2
Sample
Input
5
Output
*
***
*****
*******
*********
*******
*****
***
*
#include<stdio.h>
int main()
{
int n,i,t,t2,clock;
scanf("%d",&n);
t=n;
t2=1;
clock=2*n-1;
while (clock!=n-1)
{
for (i=t-1;i>0; i--)
{
printf(" ");
}
for (i=t2; i>0; i--)
{
printf("*");
}
printf("\n");
t=t-1;
t2=t2+2;
clock--;
}
t=2*n-2;
t2=1;
while (clock!=0)
{
for (i=t2; i>0; i--)
{
printf(" ");
}
for (i=t;i!=0; i--)
{
printf("*");
}
printf("\n");
t=t-1;
t2++;
clock--;
}
return 0;
}
*/
#include<stdio.h>
int main()
{
int n,i,j;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=n-i;j++)
printf(" ");
for(j=1;j<=i;j++)
printf("*");
for(j=1;j<i;j++)
printf("*");
printf("\n");
}
for(i=n-1;i>0;i--)
{
for(j=1;j<=n-i;j++)
printf(" ");
for(j=1;j<=i;j++)
printf("*");
for(j=i-1;j>0;j--)
printf("*");
printf("\n");
}
return 0;
}

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/*
Description
153=13+53+33
m和n范围内的水仙花数
Input
m和n100<=m<=n<=999
Output
m,n;
no;
Sample
Input
100 120
300 380
Output
no
370 371
*/
#include<stdio.h>
int main ()
{
int m,n,n1,n2,t,i,sum=0,judge=0,t2,i2;
while (scanf("%d %d",&m, &n)!=EOF)
{
t2=0;
if(m>n)
{
t=n;
n=m;
m=t;
}
for (i=m; i<=n; i++)//求范围内水仙花数
{
for (i2=3,t=i; i2>0; i2--)//循环取数求立方和
{
n1=t%10;
n2=n1*n1*n1;
sum=sum+n2;
t=t/10;
}
if (sum==i)
{
if (t2==0)
{
printf("%d",i);//输出第一个水仙花数
}
else
{
printf(" %d",i);//输出非第一个数
}
t2++;
judge++;
sum=0;//sum初始化
}
else
{
sum=0;
}
}
if (judge==0)
{
printf("no");
}
printf("\n");//换行
}
return 0;
}

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#include <stdio.h>
int main()
{
double x, sum, t, item;
int n, i;
while (scanf("%lf %d", &x, &n)==2)
{
t = x*x;
item = 1.0;
sum = 1.0;
for (i=1; i<=n; i++)
{
item *= -t;
item /= (i*2-1)*(i*2);
sum += item;
}
printf("%.4lf\n", sum);
}
return 0;
}

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/*
Description
11
Input
n (1 < n <= 10^6)
Output
n YESNO()
Sample
Input
11
13
Output
YES
NO
*/
#include<stdio.h>
int main()
{
long long int n,temp = 0,sum=0,i;
while (scanf("%lld",&n)!=EOF)
{
if(n==1||n==2)
{
printf("YES\n");
continue;
}
temp=0;
sum=0;
if(n>10)
{
temp=n;
while (temp>0)
{
sum=sum+temp%10;
temp=temp/10;
}
}
else
sum=n;
for (i=2; i<n; i++)
{
if(n%i==0)
temp++;
else
continue;
}
for (i=2; i<sum; i++)
{
if(sum%i==0)
temp++;
else
continue;
}
if (temp==0)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
return 0;
}

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/*
Description
Your task is to Calculate the sum of some integers.
Input
Input contains multiple test cases. Each test case contains a integer N, and then N integers follow in the same line. A test case starting with 0 terminates the input and this test case is not to be processed.
Output
For each group of input integers you should output their sum in one line, and with one line of output for each line in input.
Sample
Input
4 1 2 3 3
5 1 2 4 4 5
0
Output
9
16
*/
#include <stdio.h>
int main()
{
int n,i,sum,x;
while(scanf("%d",&n)!=EOF&&n)
{
sum=0;
for(i=0; i<n; i++)
{
scanf("%d",&x);
sum=sum+x;
}
printf("%d\n",sum);
}
return 0;
}

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/*
Description
C语言的学习是我们程序设计基础的重点和主要内容
(a1s4z5)饿饿
112233...n次的时候掰n个n次的时候就回家吃玉米
m次掰的玉米全都扔掉才能回家开饭(li)(hai)
Input
n和m(0 < m < n < 10^4)
Output
Sample
Input
5 2
6 3
Output
12
15
*/
#include<stdio.h>
int main()
{
int a,sum=0,i,n,m,sum2=0;
while (scanf("%d %d",&n,&m)!=EOF)
{
sum=0;
sum2=0;
for (i = 0; i <= n; i++)
{
sum = sum + i;
}
for (i = 0; i <= m; i++)
{
sum2 = sum2 + i;
}
printf("%d\n", sum - sum2);
}
return 0;
}

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#include<stdio.h>
int main()
{
int n,flag,i,i2,count = 0;
scanf("%d",&n);
for (i=2; i<n; i++)
{
flag=0;
if (n==2)
{
printf("2 ");
}
for (i2=2; i2<i; i2++)
{
if (i%i2==0)
{
flag++;
break;
}
else
continue;
}
if (flag==0)
{
printf("%d ",i);
count++;
}
if (count%10==0)
{
printf("\n");
}
}
return 0;
}

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/*
Description
, n n n 9
Input
EOF n (0 < n < 10)
Output
n
Sample
Input
2
3
Output
1*1=1
1*2=2 2*2=4
1*1=1
1*2=2 2*2=4
1*3=3 2*3=6 3*3=9
Hint
使for循环
if(n == 1) printf(1*1=1\\n);
if(n == 2) printf(1*1=1\\n1*2=2 2*2=4\\n);
SubmitSolutions
*/
#include<stdio.h>
int main()
{
int n,i,i2,mu,flag=0;
while (scanf("%d",&n)!=EOF)
for (i=1;i<=n;i++)
{
flag=0;
for (i2=1;i2<=i;i2++)
{
mu=i*i2;
if (flag==0)
printf("%d*%d=%d",i2,i,mu);
else
printf(" %d*%d=%d",i2,i,mu);
flag++;
}
printf("\n");
}
return 0;
}

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/*
Description
X晚上睡不着的时候不喜欢玩手机
Input
N(N <= 10000)Bx,yT(x,y)N个XY
Output
Sample
Input
3
0 1
3 4
1 1
2 2
3 3
2
1 1
5 5
4 4
0 6
Output
1
1
*/
#include <stdio.h>
int main()
{
int n,sum,i,x2,y2,x0,y0,x1,y1,temp;
while (scanf("%d",&n)!=EOF)
{
scanf("%d %d",&x0,&y0);
scanf("%d %d",&x1,&y1);
if(x0>x1)
{
temp=x0;
x0=x1;
x1=temp;
}
if(y0>y1)
{
temp=y1;
y0=y1;
y1=temp;
}
sum=0;
for(i=0;i<n;i++)
{
scanf("%d %d",&x2,&y2);
if ((x2<x1&&x2>x0)&&(y2<y1&&y2>y0))
{
sum++;
}
}
printf("%d\n",sum);
}
return 0;
}

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/*
Description
n的整数序列
Input
n1n10
n个正整数组成的序列
Output
Sample
Input
6
2 3 8 1 4 5
Output
1 3 5 2 4 8
*/
#include<stdio.h>
int main()
{
int n,i,t=0,t2=65535,temp,ima=0,imi=0;
scanf("%d",&n);
int a[n];
for(i=0;i<n;i++)
{
scanf("%d", &a[i]);
}//输入
for(i=0;i<n;i++)//选出最大最小值
{
if(a[i]>=t)//最大
{
t=a[i];
ima=i;
}
if(a[i]<=t2)//最小
{
t2=a[i];
imi=i;
}
}
temp=a[0];
a[0]=a[imi];
a[imi]=temp;
temp=a[n-1];
a[n-1]=a[ima];
a[ima]=temp;
for (i=0;i<n;i++)//输出
{
if(i==0)
printf("%d",a[0]);
else
printf(" %d", a[i]);
}
return 0;
}

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/*
Description
5
1
2
3
Input
5
Output
3
Sample
Input
123
Output
3
1 2 3
3 2 1
*/
#include<stdio.h>
int main()
{
long int a,n;
int i=0,count=0;
scanf("%ld",&a);
n=a;
while (n>0)
{
n=n/10;
count++;
}
int num[count];
n=a;
while (n>0)
{
num[i]=n%10;
i++;
n=n/10;
}
printf("%d\n",count);
for(i=count-1;i>=0;i--)
{
if (i==count-1)
printf("%d",num[count-1]);
else
printf(" %d",num[i]);
}
printf("\n");
for(i=0;i<count;i++)
{
if (i==0)
printf("%d",num[0]);
else
printf(" %d",num[i]);
}
printf("\n");
return 0;
}

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/*
Description
Input
N(1 <= N <= 10)
N个整数M(0 <= M <= 100)
Output
Sample
Input
3
1 2 3
5
2 4 4 5 5
Output
1 1 1
1 2 2
*/
#include<stdio.h>
int main()
{
int nmax,nmin,ave,n,i,sum,count;
while (scanf("%d",&n)!=EOF)
{
nmax = 0;
nmin = 0;
sum=0;
count=0;
int a[n];
for (i = 0; i < n; i++)
{
scanf("%d", &a[i]);
sum = sum + a[i];
}
ave = sum / n;
for (i = 0; i < n; i++)
{
if (a[i] > ave)
nmax++;
if (a[i]==ave)
count++;
if (a[i]<ave)
nmin++;
}
printf("%d %d %d\n",nmin,count,nmax);
}
return 0;
}

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/*
Description
n A1A2 An LR[LR]()
Input
n 1< n < 10000
n int
LR0 < L < R <= n
Output
[LR] int
Sample
Input
5
3 5 6 2 9
2 4
Output
13
*/
#include<stdio.h>
int main()
{
int n=0, sum=0, l, r, i;
scanf("%d", &n);
int a[n];
for (i = 0; i < n; i++)
{
scanf("%d", &a[i]);
}
scanf("%d %d", &l, &r);
for (i = l-1; i < r; i++)
{
sum = sum + a[i];
}
printf("%d\n",sum);
return 0;
}

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/*
Description
10
Input
n
Output
Sample
Input
678123
Output
6 7 8 1 2 3
*/
#include<stdio.h>
int main()
{
long long int num,sum=0,n,i=0;
scanf("%lld",&num);
n=num;
while (n>0)
{
n=n/10;
sum++;
}
int a[sum];
for(i=0;i<sum;i++)
{
a[i]=num%10;
num=num/10;
}
for(i=sum-1;i>=0;i--)
{
if (i==sum-1)
{
printf("%d",a[sum-1]);
}
else
{
printf(" %d",a[i]);
}
}
printf("\n");
return 0;
}

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/*
Description
n A1A2 An LR[LR]()
Input
n 1< n < 10000
n int
LR0 < L < R <= n
Output
[LR] int
Sample
Input
5
3 5 6 2 9
2 4
Output
13
*/
/*TLE
#include<stdio.h>
int main()
{
int n,i,i2,count=0,judge=50;
while ((scanf("%d",&n)!=EOF)||(judge!=0))
{
int a[n], c[n];
for (i = 0; i < n; i++)
{
scanf("%d", &a[i]);
}
for (i = 0; i < n; i++)
{
for (i2 = 0; i2 < n; i2++)
{
if (a[i] == a[i2])
{
count++;
}
}
c[i] = count;
count = 0;
}
i2 = 0;
for (i = 1; i < n; i++)
{
if (c[i - 1] > c[i])
{
i2 = i-1;
}
else
{
i2 = i;
}
}
printf("%d\n", a[i2]);
judge--;
}
return 0;
}
*/
#include <stdio.h>
int main()
{
int a,n;
int i,k;
while(~scanf("%d",&n))
{
int b[1001]={0},max=0;
for(i=0;i<n;i++)
{
scanf("%d",&a);
b[a]++;
/*统计“0”到“1000”对应的数出现的次数例如若输入54便在“b[54]”中加“1”
b[54]=354*/
}
for(i=0;i<=1000;i++) //寻找最大次数;
{
if(b[i]>max)
{
max = b[i];
k = i; //标记其下标;
}
}
printf("%d\n",k);
}
return 0;
}

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/*
Description
110101线
Input
10
1010110m1N号的N个选手的成绩m,m范围是(0 < m < 100)
Output
Sample
Input
10
2 5 3 9 7 10 23 12 43 5
10
6 1 7 9 3 4 8 3 2 9
Output
1
6
*/
#include<stdio.h>
int main()
{
int n, a[10], i, i2, num = 1;
while (scanf("%d", &n) != EOF)
{
for (i = 0; i < 10; i++)
{
scanf("%d", &a[i]);
}
for (i = 1; i < 10; i++)
{
if (a[0] > a[i])
{
num++;
}
}
printf("%d\n", num);
num = 1;
}
return 0;
}

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/*
Description
AC^_^
Input
n<100n行是n组数据3
Output
Sample
Input
2
2009 1 1
2008 1 3
Output
1
3
*/
#include<stdio.h>
int main()
{
int i,n;
int y,m,d,sum,i2;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
{
int c[12]={31,28,31,30,31,30,31,31,30,31,30,31};
sum=0;
scanf("%d %d %d",&y,&m,&d);
if ((y% 4 == 0 && y % 100 != 0) || y% 400 == 0)
{
c[1]=29;
}
if (m!=1)
{
for (i2=1; i2<m; i2++)
{
sum=sum+c[i2-1];
}
}
sum=sum+d;
printf("%d\n",sum);
}
}
return 0;
}

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@ -0,0 +1,55 @@
#include<stdio.h>
#include<string.h>
int main()
{
long int j,n,len ;
char str[100],t,le[5];
scanf("%ld",&n);
while(n>0)
{
memset(str,0,sizeof(str));
memset(le,0,sizeof(le));
gets(str);
len = strlen(str);
for(j=0;j<len;j++)
{
if(t>='A'&&t<='Z')
str[j]=str[j]+32;
t=str[j];
switch (t)
{
case 'a':
le[0]++;
break;
case 'e':
le[1]++;
break;
case 'i':
le[2]++;
break;
case 'o':
le[3]++;
break;
case 'u':
le[4]++;
break;
}
}
printf("a:%d\n",le[0]);
printf("e:%d\n",le[1]);
printf("i:%d\n",le[2]);
printf("o:%d\n",le[3]);
printf("u:%d\n",le[4]);
n--;
}
return 0;
}

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#include<stdio.h>
int main ()
{
int n,i,a,sum=0,i2;
while(scanf("%d",&n)!=EOF)
{
int a[n];
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
if(((i>1)||((i+1)<n)))
{
if((a[i]>a[i+1])&&(a[i]>a[i-1]))
{
sum++;
}
}
}
printf("%d\n",sum);
sum=0;
}
return 0;
}

View File

@ -0,0 +1,110 @@
/*
Description
N(N<=100),
Input
N,N个整数
Output
N个数
Sample
Input
5
1 4 3 2 5
Output
1 2 3 4 5
*/
#include<stdio.h>
int main()
{
int n,i,t;
scanf("%d",&n);
int a[n];
for (i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
for (i=0; i<n-1; i++)
{
if(a[i] > a[i+1])
{
t = a[i];
a[i] = a[i+1];
a[i+1] = t;
i=0;
continue;
}
}
if(a[0] > a[1])
{
t = a[0];
a[0] = a[1];
a[1] = t;
}
for (i=0; i<n; i++)
{
if (i==0)
{
printf("%d",a[0]);
}
else
printf(" %d",a[i]);
}
printf("\n");
return 0;
}
#include<stdio.h>
int main()
{
int n,i,t,j;
scanf("%d",&n);
int a[n];
for (i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
for (i=0; i<n-1; i++)
{
for(j = 0; j < n - 1 - i; j++)
{
if(a[j] > a[j+1])
{
t = a[j];
a[j] = a[j+1];
a[j+1] = t;
}
}
}
for (i=0; i<n; i++)
{
if (i==0)
{
printf("%d",a[0]);
}
else
printf(" %d",a[i]);
}
printf("\n");
return 0;
}

View File

@ -0,0 +1,62 @@
/*
Description
N(N<=100),
Input
N,N个整数
Output
N个数
Sample
Input
5
1 4 3 2 5
Output
1 2 3 4 5
*/
#include<stdio.h>
int main()
{
int n,i,t,j;
scanf("%d",&n);
int a[n];
for (i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
for (i=0; i<n-1; i++)
{
for(j = 0; j < n - 1 - i; j++)
{
if(a[j] > a[j+1])
{
t = a[j];
a[j] = a[j+1];
a[j+1] = t;
}
}
}
for (i=0; i<n; i++)
{
if (i==0)
{
printf("%d",a[0]);
}
else
printf(" %d",a[i]);
}
printf("\n");
return 0;
}

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