1001 Sum Problem

Sum Problem

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 387857    Accepted Submission(s): 97127

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

1 100

 

 

Sample Output

1 5050

 

 

#include <iostream>

using namespace std;

int main()
{
    int n,sum;
    while(cin>>n)
    {
        if(n%2!=0)
        sum=(1+n)/2*n;
        else
        sum=(1+n)/2*n+n/2;
        cout<<sum<<endl;
        cout<<endl;
    }
    return 0;
}

注意！！注意！！！

以为是一个很简单的问题，但会不小心的把（You may assume the result will be in the range of 32-bit signed integer.）给忽略掉，当n*(n-1)时是会溢出的。