NYOJ 26 A problem is easy
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 ?
Input
The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10[sup]10[/sup]).
Output
For each case, output the number of ways in one line.
Sample Input
2
1
3
Sample Output
0
1
题目大意：
给出一个数n，求 i * j + i + j = n(0 < i <= j)有多少种方法。
#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
int main()
{
long long int a,b,c,d,e,f,m;
cin>>a;
while(a--)
{
cin>>m;
m++;
b=sqrt(m);
e=0;
for(c=2;c<=b;c++)
{
if(m%c==0&&m/c>=c)
e++;
}
cout<<e<<endl;
}
return 0;
}
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