一道简单的acm题,可是我老是通不过ojInputInputcontains multiple test cases, and one case one line. Each case starts with aninteger N, and then N integers follow in the same line. OutputFor each testcase you should output the sum of N inte

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一道简单的acm题,可是我老是通不过ojInputInputcontains multiple test cases, and one case one line. Each case starts with aninteger N, and then N integers follow in the same line. OutputFor each testcase you should output the sum of N inte

一道简单的acm题,可是我老是通不过ojInputInputcontains multiple test cases, and one case one line. Each case starts with aninteger N, and then N integers follow in the same line. OutputFor each testcase you should output the sum of N inte
一道简单的acm题,可是我老是通不过oj
Input
Inputcontains multiple test cases, and one case one line. Each case starts with aninteger N, and then N integers follow in the same line.

Output
For each testcase you should output the sum of N integers in one line, and with one line ofoutput for each line in input.
Sample Input
4 1 2 3 4
5 1 2 3 4 5


Sample Output
10
15

一道简单的acm题,可是我老是通不过ojInputInputcontains multiple test cases, and one case one line. Each case starts with aninteger N, and then N integers follow in the same line. OutputFor each testcase you should output the sum of N inte
#include<stdio.h>
 
int main()
{
    int n,x,sum;
    while(scanf("%d",&n)!=EOF)
    {
        sum = 0;
        while(n--)
        {
            scanf("%d",&x);
            sum+=x;
        }
        printf("%d/n",sum);
    }
    return 0;
}

学习一下输入输出就好,欢迎交流.

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