NYOJ 题目216 A problem is easy【推数学公式】

A problem is easy

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

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 ?

输入

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).

输出

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

样例输入

213

样例输出

01

上传者

苗栋栋

我的天，我只想到了一半，而没有转化成代码

i*j+i+j=(i+1)*(j+1)-1

       n=(i+1)*(j+1)-1

   n+1=(i+1)*(j+1)

即  n+1=i*j

只要(n+1)%i==0就可以了

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;int main(){    int t,n;   cin>>t;   while(t--)   {cin>>n;   n=n+1;   int count=0,i;   for(i=2;i*i<=n;i++)   { if(n%i==0)      count++;   }     cout<<count<<endl;   }}