ACM：Sum Problem 求和问题C语言代码_OC/C/C++

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

即：求1+2+3+…+n的和

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

C++代码


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

C语言代码


#include <stdio.h>
int main()
{
 int n,i,sum=0;
 while(scanf("%d",&n)!=EOF)
 {
  if(n==1)
  {
   sum=1;
  }
  else 
  {
   for(i=0;i<=n;i++)
   {
   sum+=i;
   }
  }
  printf("%d\n\n",sum);
  sum=0;
 }
 
 return 0;
}

C语言解法2，利用公式


#include<stdio.h>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        printf("%.f/n/n",(1+n)/2.0*n);
    }
    return 0;
}