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  • NYOJ 216:An easy problem

    Description

    When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of dough ..” , “100 yuan buy 100 pig” .etc.. 

    One day Teddy met a old man in his dream , in that dream the man whose name was“RuLai” gave Teddy a problem : 

    Given an N , can you calculate how many ways to write N as i * j + i + j (0 < i <= j) ? 

    Teddy found the answer when N was less than 10…but if N get bigger , he found it was too difficult for him to solve. 
    Well , you clever ACMers ,could you help little Teddy to solve this problem and let him have a good dream ? 

    Input

    The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10 10).

    Output

    For each case, output the number of ways in one line.

    Sample Input

    2
    1
    3

    Sample Output

    0
    

    1

    i*j+i+j=n等价于i*j+i+j+1=n+1

    分离变量得:(i+1)*(j+1)=n+1

    也就是i*j=n+1

    i从2开始到sqrt(n+1);

    #include<cstdio>
    #include<cmath>
    int main()
    {
    	__int64 t;
    	while(scanf("%d",&t)!=EOF)
     {
    	while(t--)
    	{
    		__int64 n,cnt=0;
    		scanf("%I64d",&n);
    		for(int i=2;i<=sqrt(n+1);i++)
    		{
    			if((n+1)%i==0)
    			{
    				cnt++; 
    			}
    		}
    		printf("%I64d
    ",cnt);
    	}
     }
    
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027021.html
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