hdu 1852(快速幂模+有除法的时候取模的公式)

Posted 2020-07-11 AC菜鸟机

Beijing 2008

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 741    Accepted Submission(s): 291

Problem Description

As we all know, the next Olympic Games will be held in Beijing in 2008. So the year 2008 seems a little special somehow. You are looking forward to it, too, aren‘t you? Unfortunately there still are months to go. Take it easy. Luckily you meet me. I have a problem for you to solve. Enjoy your time.

Now given a positive integer N, get the sum S of all positive integer divisors of 2008^N. Oh no, the result may be much larger than you can think. But it is OK to determine the rest of the division of S by K. The result is kept as M.

Pay attention! M is not the answer we want. If you can get 2008^M, that will be wonderful. If it is larger than K, leave it modulo K to the output. See the example for N = 1，K = 10000: The positive integer divisors of 20081 are 1、2、4、8、251、502、1004、2008，S = 3780, M = 3780, 2008^M % K = 5776.

 

 

Input

The input consists of several test cases. Each test case contains a line with two integers N and K (1 ≤ N ≤ 10000000, 500 ≤ K ≤ 10000). N = K = 0 ends the input file and should not be processed.

 

 

Output

For each test case, in a separate line, please output the result.

 

 

Sample Input

1 10000 0 0

 

 

Sample Output

5776

 

收获挺大的！。以前对于除法模运算只知道用逆元可以算，，但是当两个数不互素的时候就不知道怎么弄了。今天得到了两个公式。。第一个公式自己做的时候想到了可能可以，然后真的AC了，然后去验证发现真的有：

1.(a/b)%mod=a%(b*mod)/b%mod;(get这个公式好激动)

2.(a/b)%mod=a*b^(mod-2)%mod，mod为素数（可以通过逆元证明）（这个公式的话感觉如果mod为素数的话，直接用逆元也一样的，，可以参考我博客hdu1452）

然后这个题并不难，把2008分解成 251*2^3 然后求因子和用第一个公式去掉分母250,然后可以得到M，在用快速幂计算就好了。

 

#include <stdio.h>
#include <iostream>
using namespace std;
typedef long long LL;

LL pow_mod(LL a,LL n,LL mod){
    LL ans = 1;
    while(n){
        if(n&1) ans = a*ans%mod;
        a=a*a%mod;
        n=n>>1;
    }
    return ans;
}

int main()
{
    LL N,K;
    while(scanf("%lld%lld",&N,&K)!=EOF,N&&K)
    {
        K = 250*K;
        LL M = ((pow_mod(251,N+1,K)-1)*(pow_mod(2,3*N+1,K)-1))%K;
        M = M/250;
        K/=250;
        LL ans =pow_mod(2008,M,K);
        printf("%lld\n",ans);
    }
    return 0;
}