数学 hdu 2601 （An easy problem）

An easy problem

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 <= 10¹⁰).

 

Output

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

 

Sample Input

213

 

Sample Output

01

 题意：问n=i*j+i+j有多少组合方式

题解：n+1=(i+1)*(j+1) ,  因为i<=j, i+1<=sqrt(n+1) , j+1=(n+1)%(i+1).

代码：

#include<cstdio>#include<cmath>using namespace std;int main(){int t;long long n,i,j,ans;scanf ("%d",&t);while (t--){ans=0;scanf ("%lld",&n);n+=1;for (i=2;i<=sqrt(n);i++){if (n%i==0)ans++;}printf ("%lld\n",ans);}return 0;}