(解题报告）HDU1001---Sum problem

Sum Problem

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

Total Submission(s): 372618 Accepted Submission(s): 93378

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

解题要点

用while(scanf()!=EOF)输入；

两种求法，第一种直接利用for累加，第二种利用公式求；

第一种

注意数据范围，因使用long long 类型———定义 __int64 调用 %I64d

第二种

使用sum=n*(n+1)/2错误，原因是n * (n+1)的数据范围可能超过32位，所以应该使用sum=n/2 * （n+1）或 sum=(n+1)/2 * n;

注意每个例子后面都有一行空行

第一种代码如下：

#include <stdio.h>
int main ()
{
int i,n;
__int64 sum;
while(scanf("%d",&n)!=EOF)
{
sum=0;          //每算完一次都将sum清0
for(i=1;i<=n;i++)
{
sum+=i;
}
printf("%I64d\n",sum);
printf("\n");
}
return 0;
}

第二种代码如下:

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