杭电oj1001（c++）

题目分析

Problem Description

Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).

In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + … + n.

Input

The input will consist of a series of integers n, one integer per line.

Output

For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.

题目要求的输入输出样式为先输入数字n，然后输出sum(n)独占一行，后面紧跟一空行。由于"for each case"，该题也是循环过程。

代码

#include 
using namespace std;
int main()
{
	int a, b, i;
	while(cin >> a )
	{
		b=0;
		for(i=1;i<=a;i++){
			b+=i;
		}
		cout << b << endl;
		cout << endl;
	}
	return 0;
}

注意事项

该题不可用高斯算法n*(n+1)/2.
You may assume the result will be in the range of 32-bit signed integer.
要求结果在32位有符号整型范围内(int)。即便你的最终结果n*(n+1)/2在32位内，n*(n+1)也有可能溢出这个范围。