Problem 1001

Sum Problem

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 184771    Accepted Submission(s): 43962

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.

 

Sample Input

1100

 

Sample Output

15050

 

Author

DOOM III

因为n很大的时候n*(n+1)的值会溢出。可以判断奇偶然后先做除法。注意这句：You may assume the result will be in the range of 32-bit signed integer.要求是你的结果要在32位有符号整型范围内。假设n*(n+1)/2的结果刚好满足32位。但是在做除法前n*(n+1)的结果就超过了32位。

来自：http://zhidao.baidu.com/question/502897991.html&from=emailmsg#submit

感谢 DN_海伦泰勒 

#include <iostream>using namespace std;void main(){int n;while(cin>>n) {int i,sum=0;for(i=1;i<=n;i++) sum=sum+i;cout<<sum<<endl<<endl;}}

#include <iostream>using namespace std;void main(){int n;while(cin>>n) {cout<<n*(n+1)/2<<endl<<endl;}}