杭电ACM1001

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

其实博主刚看见这题的时候真的是一脸懵逼，后来....(⊙o⊙)…百度了答案.......好像就是一个前n项求和

当然啦，这都只是说的容易做着难呐，读懂题，有想法才是关键

#include<iostream>

using namespace std;

int main(){

int n;

while(cin>>n){

if(n%2==1)

{

cout<<(1+a)/2*a<<endl<<endl;

}

else

{

cout<<a/2*(a+1)<<endl<<endl;

}

}

return 0;

}