题干：

Beijing 2008

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 917    Accepted Submission(s): 394

 

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 2008N. 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 2008M, 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, 2008M % 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

 

 

Author

lxlcrystal@TJU

 

题意：求2008^n的所有因子和m对k取余，然后求2008^m对k取余。

解题报告：

这题是类似的一题，可以除法可以通过求逆元。因为那题的mod是29，gcd(2*166,29)==1，存在逆元。但是这里的gcd(250,k)不一定满足等于1，也就是说不一定存在逆元。那么观察我们要求的m，m肯定有250这个分母，所以可以表示成m=x/250。m%k=(x/250)%k转化为(x%(250*k))/250。求变成了先除法再取余了。

 

AC代码：

#include<bits/stdc++.h>
#define LL long long
using namespace std;

LL q_pow(LL a,LL k,LL mod){
	LL ans=1;
	while(k){
		if(k&1){
			ans=(ans*a)%mod;
		}
		k>>=1;
		a=(a*a)%mod;
	}
	return ans;
}
// 那么观察我们要求的m，m肯定有250这个分母，
// 所以可以表示成m=x/250。m%k=(x/250)%k转化为(x%(250*k))/250。求变成了先除法再取余了。
int main()
{
	int n,k;
	LL a,b,ans;
	while(~scanf("%d%d",&n,&k) ) {
		if(n==0&&k==0) break;
		a=q_pow(2,3*n+1,250*k)-1;
		b=q_pow(251,n+1,250*k)-1;//mod取250*k，目的是先要保留250这个因子，放在取余后再除
		ans=(a*b)%(250*k);
		ans/=250;
		ans=((ans%k)+k)%k;
		printf("%lld\n",q_pow(2008,ans,k));
		
	}
	
	return 0 ;
}

类似的一题：

https://blog.csdn.net/qq_41289920/article/details/81190874