A problem is easy

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

 #include <iostream>#include <cmath>using namespace std;int main(){    int n;    int a,b;    cin>>n;    while(n--)    {        cin>>a;        b=0;        for(int i=2;i*i<=(a+1);i++)        {            int j = (a+1)/i;            if(i*j==a+1 && i<=j)                        b++;        }        cout<<b<<endl;    }return 0;}        

 

A problem is easy