HDU-1001 Sum Problem

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

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

 

Author

DOOM III

正确解

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

注

一、题目求1+2+3+......+n的值

二、输出结果后要有空白行

错误解

#include <stdio.h>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
        printf("%d\n\n",n*(n+1)/2);
    return 0;
} 

错误原因

因为求和公式我们最后会除以2，题目中指明了result是在32位范围内，但并不代表在除以2之前，也就是运行到(n*(n+1))这里，值还是在范围里面