An easy problem

An easy problem Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit ​​Status​​

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

#include<iostream> #include<stdio.h> #include<math.h> using namespace std; int main() {     int t;     cin>>t;     while(t--)     {         long long n;         scanf("%I64d",&n);         n++;         int ans=0;         long long m=sqrt(n);         for(long long i=2;i<=m;i++)         {             if(n%i==0) ans++;         }         printf("%d\n",ans);     } }

View Code

n=i*j+i+j

==>n+1=(i+1)*(j+1)