基础数论 Beijing 2008 HDU - 1852 （快速幂）

最新推荐文章于 2020-05-15 15:28:43 发布

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最新推荐文章于 2020-05-15 15:28:43 发布

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本文介绍了一个算法问题的解决方案，该问题要求计算2008^n的所有正整数因子之和，并进一步计算特定模数下的幂运算结果。通过分解质因数和利用等比数列求和公式，文章提供了高效的计算方法。

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Beijing 2008

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

题意：给定N和K，
首先求S，S为2008^N所有因子之和，分解2008^N=（2^3*251）^N=(2^3N )*(251^N)
2^3N , 因数有2^0 , 2^1 , 2^2,,,,,,,,2^3N
251^N ,因数有251^0,251^1,,,,251^N
令A=2^0 +2^1 + 2^2+,,,,,,,+2^3N
B=251^0+251^1+,,,,+251^N
所有因子之和=A*B=S
等比求和得到A=2^(3*N+1)-1;
B=(251^(N+1)-1)/250;
直接求S肯定超long long范围
然后求M，M=S mod K;
因此在求S时取模，在快速幂中取模
最后求2008^M%K

//快速幂
ll pow(ll a,ll m,ll mod)//分别是底数，幂，模
{   
        int res = 1;
        while(m){
            if(m&1){
                res=res*a%mod;
            }
            a=a*a%mod;
            m>>=1;
        }
        return res;
}

另外在求B时：因为B=[（251^(N+1)-1）/250 ]mod K
把250拿出来 B=(251^(N+1)-1)mod(250*K)/250;

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#define ll long long
using namespace std;
ll n,k;
ll pow(ll a,ll m,ll mod)
{
    ll res=1;
    while(m){
        if(m&1){
            res=(res*a)%mod;
        }
        a=a*a%mod;
        m>>=1;
    }
    return res;
}
int main()
{
    while(scanf("%lld %lld",&n,&k)){

        if(n==0&&k==0) break;
        ll M=250*k;
        ll A=pow(2,3*n+1,M)-1;
       // cout<<A<< endl;
        ll B=pow(251,n+1,M)-1;
        //cout << B << endl;
        ll m=(A*B)%M/250;
    //printf("----%lld\n",m);
        ll s=pow(2008,m,k);
        printf("%lld\n",s);

    }
    return 0;
}