hdu 2619 Love you Ten thousand years（数论，待解决）

最新推荐文章于 2020-12-27 19:27:03 发布

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探讨了在特定数学条件下，如何计算一个基于n的LovePoint的数量，这被比喻为地球旋转代表的爱的天数。输入一个大于等于3的整数n，通过算法找出所有可能的LovePoint，并解释了一种高效的求解方法。

Love you Ten thousand years

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 92 Accepted Submission(s): 34

Problem Description

Love you Ten thousand years------Earth's rotation is a day that is the representative of a day I love you. True love, there is no limit and no defects. Earth's revolution once a year, it is on behalf of my love you more than a year. Permanent horizon, and my heart will never change ……

We say that integer x, 0 < x < n,(n is a odd prime number) is a LovePoint-based-on n if and only if the set { (xⁱ mod n) | 1 <= i <= n-1 } is equal to { 1, ..., n-1 }. For example, the powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus 3 is a LovePoint-based-on 7.
Now give you a integer n >= 3(n will not exceed 2³¹).
We say the number of LovePoint-based-on n is the number of days the earth rotating.
Your task is to calculate the number of days someone loved you.

Input

Each line of the input contains an integer n. Input is terminated by the end-of-file.

Output

For each n, print a single number that gives the number of days someone loved you.

Sample Input

Sample Output

看了网上的一份题解，对n做两次除以所有质因数乘以质因数-1即可得到答案，为什么？？

完全不是夜深人静写算法里面说的找规律，后来人如果发现另外的解法或者此种解法的缘由，还望指点。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define LL long long
int vis[100];
LL get(LL n)
{
    LL ans=n;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            ans=ans/i*(i-1);
            while(n%i==0)
                n/=i;
        }
    }
    if(n>1) ans=ans/n*(n-1);
    return ans;
}
int main()
{
    LL m;
    while(scanf("%lld",&m)!=EOF)
    {
        printf("%lld\n",get(get(m)));
    }
    return 0;
}