1001 Sum Problem

Sum Problem

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

Total Submission(s): 341647    Accepted Submission(s): 86129

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

1100

 

Sample Output

15050

这个题是求自然数前n项的和，结果在int 范围内，有两种处理方法，一个是循环累加，一个是求和公式，下面我用的是求和公式.....

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

提交上去，wrong answer！！

为什么，想来想去，才发现这个程序的漏洞，你想到了吗？输个五万（50000）上去试试就知道了......

就是说，数据结果不超过int 但是中间过程可能超过int ，这里有两种方法解决

1，用long long 型运算，2，用另外一种方法.......

这样的方法可以防溢出，不懂的话，好好想想为什么，想通了的确对自己有帮助，这是个很好的思想..嘿嘿

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