HDU 2601 An easy problem(求因数)

博客介绍了一个有趣的数学问题,即给定一个整数N,如何找出所有可能的i和j (0<i≤j),使得i*j+i+j等于N。通过巧妙的转换,问题变为寻找N+1的所有非1非自身的因数。

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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

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;
}



 
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