HDU1825【唯一分解定理+积性函数】-优快云博客

Discription
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
现在给定一个正整数N，获得2008^N的所有正整数除数的总和S模k的值为M。
求解2008^M。

思路
这道题跟HDU 1452：Happy 2004很像。但是250和2008不互质，不能求逆元，想了很久。问了下ZJH才反应过来可以不用求逆元。

AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll qmi(ll m, ll k, ll p)
{
    ll res=1%p,t=m;
    while(k)
    {
        if (k&1)
            res=res*t%p;
        t=t*t%p;
        k>>= 1;
        //cout<<res<<endl;
    }
    //cout<<res<<endl;
    return res;
}

ll n,kk;
ll ans,tmp,a,b;

int main()
{
    while(scanf("%lld%lld",&n,&kk))
    {
        if(n==0&&kk==0)
            break;
        tmp=(qmi(2,3*n+1,250*kk)-1)*(qmi(251,n+1,250*kk)-1)%(250*kk)/250;
        //cout<<a<<" "<<b<<endl;
        //cout<<tmp<<endl;
        ans=qmi(2008,tmp,kk);
        cout<<ans<<endl;
        //qmi(250,kk-2,kk);
    }
    return 0;
}