HDU 2601 An easy problem（求因数）

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 ?

 

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

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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;
}