一道简单的acm题,可是我老是通不过oj
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
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
赞 lingling77 幼苗
共回答了21个问题采纳率：90.5% 举报
#include
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.
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.
1年前 追问
1
给个题目网址吧，我试一下。
#include
int main()
{
int n,x,sum;
while(scanf("%d",&n)!=EOF)
{
sum = 0;
while(n--)
{
scanf("%d",&x);
sum+=x;
}
printf("%dn",sum);
}
return 0;
}
这个是可以AC的，你试试哈。
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你能帮帮他们吗