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

样例输出

0
1

程序主要目的是怎么简化时间复杂度，想到原式等价于（m+1）=（i+1）*（j+1），所以问题就归结于求（m+1）大于等于2的因子的个数.

#include <stdio.h>
#include <math.h>

int main(){
    int T;
    int n;
    int amount;
    int i;

    
    scanf("%d",&T);
    
    while(T--){
        scanf("%d",&n);
        
        amount=0;
        for(i=2;i<=sqrt(n+1);i++){
            if((n+1)%i==0)
                amount++;
        }
        
        
        printf("%d
",amount);
    }
    
    return 0;
}