hdu 1852 Beijing 2008（快速幂取模）

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本文介绍了一种高效的算法来解决特定形式的大数模运算问题，即计算2008的n次方对某个较大数k取模的结果，并进一步计算2008对该结果取模的值。通过使用快速幂运算技巧和巧妙的数学转换，避免了直接计算带来的巨大开销。

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Beijing 2008

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 744 Accepted Submission(s): 294

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

题意：给你n和k，2008的n次方对k取余为m，求2008的m次方对k取余

思路：不要用库函数的pow很慢，用快速幂，只要log（n）的时间。

另外，这道题不能用乘法逆元去处理k，因为乘法逆元要求必须互素。

所以我们可以运用这个公式

x/d%k==x%(d*k)/d这样就可以直接得到答案了。注意取模就要全部都对(d*k)取了~

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define LL long long
LL pow_mod(LL a,LL n,int k)
{
    LL ans=1;
    while(n)
    {
        if(n&1) ans=ans*a%k;
        a=a*a%k;
        n>>=1;
    }
    return ans;
}
int main()
{
    int n,k;
    while(scanf("%d %d",&n,&k)&&(n+k))
    {
        LL ans=1;
        LL temp=pow_mod(2,3*n+1,250*k)-1;
        LL temp1=pow_mod(251,n+1,250*k)-1;
        ans=temp*temp1%(250*k)/250;
        ans=pow_mod(2008,ans,k);
        printf("%lld\n",ans);
    }
    return 0;
}