A problem is easy

A problem is easy
时间限制：1000 ms | 内存限制：65535 KB
难度：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 ?
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
问题连接：http://acm.nyist.net/JudgeOnline/problem.php?pid=216

问题分析：
转换 i * j + i + j = n 为(i +1)*(j +1) = n+1，此时只要找到(n+1)%i == 0 的i即可，以此就对i、j的寻找变为对i的寻找。从而减小搜索的时间复杂度。
0 < i <= j

#include <iostream>#include <stdio.h>#include <math.h>/* run this program using the console pauser or add your own getch, system("pause") or input loop */int main(int argc, char** argv) {    int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        // i * j + i + j   ==>(i +1)*(j +1) = n+1   0 < i <= j        int sum = 0;        for(int i=2;i*i<=(n+1);i++){            if(((n+1)/i > 1) && (n+1)%i == 0){                sum++;            }        }        printf("%d\n",sum);    }    return 0;}