杭电acm1001 Snm Problem的解法

Sum Problem

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 319745    Accepted Submission(s): 80523

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
 
由于有be in the range of 32-bit signed integer的限制，我刚开始做的 n*(n+1)/2通不过，打个比方，就像给出的结果必须在100以内，你算出的数为90，但是在没有除2的时候就超出范围了。
 
正确的代码：
#include<iostream> 
using namespace std; 
int main() 
{  
    int a;  
    while(cin>>a) 
    {  
        if(a%2 == 1)  
            cout<<(1+a)/2*a<<endl<<endl; 
        else 
        {  
            cout<<a/2*(a+1)<<endl<<endl; 
        } 
    }  
    return 0; 
}
这样就很好地解决了超限的问题，此外还有另外两种解法
1：
#include <iostream>
using namespace std;
int main()
{
    __int64 n,m;
    while(cin>>n)
    {
       m=((1+n)*n)/2;
       cout<<m;
    }
    return 0;
}
这是扩大了最大的限度。
 
2：
#include <iostream>
using namespace std;
int main()
{
    long long n,m;
    while(cin>>n)
    {
       m=((1+n)*n)/2;
       cout<<m;
    }
    return 0;
}