An easy problem
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10006 Accepted Submission(s): 2511
Problem 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 ?

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
Author
Teddy
Source
题意:
给出一个N,求出使等式 i×j+i+j=N 成立的i j的个数 (0 < i <= j)。
point:
i * j + i + j =N
i*(j+1)+j=N
i*(j+1)+j+1=N+1
(i+1)(j+1)=N+1-------X*Y=N+1
题目便转化为,求N+1的非1非自身因数有几个,且 (1 < X <= Y)
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define ll long long
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
ll n;
scanf("%lld",&n);
int ans=0;
ll m=sqrt(n+1);
for(int i=2;i<=m;i++)
{
if((n+1)%i==0)
ans++;
}
printf("%d\n",ans);
}
return 0;
}