weixin_44921406
weixin_44921406
采纳率100%
2019-09-10 22:39

求助 acm报错 改不来了 求大佬帮忙

已采纳

改了好多次了 要么PE 要么就TLE
我的代码:
#include
int main(void)
{
int M,N,i;
scanf("%d\n",&M);

while((M--)!=0)
{
while(scanf("%d ",&N),N)
{
int sum =0;
while((N--)!=0)
{
scanf("%d ",&i);
sum +=i;
}if(M !=1){
printf("%d\n\n",sum);
}
else
{ printf("%d\n",sum);
break;
}

}
}

return 0;
}

题目:
Problem Description
Your task is to calculate the sum of some integers.

Input
Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.

Output
For each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.

Sample Input
3
4 1 2 3 4
5 1 2 3 4 5
3 1 2 3

Sample Output
10

15

6

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1条回答

  • qtchen_1988 qtchen_1988 2年前
    
    #include <stdio.h>
    #include<iostream>
    #include <malloc.h>
    #include <string.h>
    
    using namespace std;
    
    int main(void)
    {
        int M,N,item,sum;
        scanf("%d",&N);
    
        int *p_sum = (int *)malloc(N*sizeof(int));
        memset(p_sum,0x0,N*sizeof(int));
        for(int i=0;i<N;i++)
        {
            sum = 0;
            scanf("%d",&M);
            while((M--)!=0)
            {
                scanf("%d",&item);
                sum += item;
            }
            *(p_sum+i) = sum;
        }
    
        printf("######Output######\n");
        for(int j=0;j<N;j++)
        {
            printf("%d\n",p_sum[j]);
        }
    
        return 0;
    }
    
    
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