HDUOJ1092
来源:互联网 发布:王者荣耀积木淘宝 编辑:程序博客网 时间:2024/06/07 03:10
HDUOJ1092
Problem 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 45 1 2 3 4 50
Sample Output
1015
My solution:
/*2015.7.20*/
#include<stdio.h>
int main()
{
int a[20],n,i,sum;
while(scanf("%d",&n)!=EOF&&n!=0)
{
sum=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
printf("%d\n",sum);
}
return 0;
}
0 0
- HDUOJ1092
- hduoj1092,A+B for Input-Output Practice (IV)
- hduoj1092(A+B for Input-Output Practice (IV))
- 【Spark Core】任务执行机制和Task源码浅析2
- HDOJ 2080 Lowest Common Multiple Plus
- polipo代理服务器简介
- AOJ AHU 173编辑距离
- MyBatis 相关下载
- HDUOJ1092
- 【Spark Core】从作业提交到任务调度完整生命周期浅析
- 关于const 作用和用法 C++ const 的全面总结
- ios修改根视图控制器
- ARC (Automatic Reference Counting)
- UVa 10142 - Australian Voting
- Git之分支创建策略
- Mysql JDBC 连接串参数说明
- C#之Lamba表达式