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

2 1 3

1 2 3

2 1 3

样例输出

0 1

1 2

0 1

[cpp]
#include “iostream”
#include “math.h”
using namespace std;
int main()
{
int i,m,n,k,t;
cin>>n;
while(n–)
{
cin>>m;
t=sqrt(m+1);
for(i=2,k=0;i<=t;i++) if((m+1)%i==0) k++; cout<