【HDU2601】An easy problem（思维）

题目链接

An easy problem Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10693    Accepted Submission(s): 2728   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 2 1 3   Sample Output 0 1   Author Teddy

【题意】

在正整数范围内，找到一个i和j满足 i * j + i + j=N，输出有多少对这样的i和j。

【解题思路】

因为i*j+i+j+1=(i+1)*(j+1) 所以（N+1）=（i+1）*(j+1)。因为i是在正整数范围内，所以i+1在>=2的范围，所以，只需查找2到sqrt(N+1)中有多少个数能使N+1被整除。

【代码】

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        LL n,ans=0;
        scanf("%lld",&n);
        for(int i=2;i<=sqrt(n+1);i++)
        {
            if((n+1)%i==0)ans++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}