ACM A problem is easy

A problem is easy

時間限制：1000 ms  |  記憶體限制：65535 KB

難度：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 ?

 

輸入

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

輸出

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

樣例輸入

2
1
3

樣例輸出

0
1

 

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main(){
    int T;
    cin >> T;
    for(int  icase  = 0 ; icase < T; icase ++){
        int n;
        cin >> n;
        n++;
        int res = 0;
        for(int i = 2; i <= (int)sqrt(n); ++ i){
            if(n%i == 0) res ++ ;
        }
        cout<< res<<endl;
    }
}