hdu1852 Beijing 2008 x^y的因子和2.0

本文链接：https://blog.youkuaiyun.com/weixin_36571742/article/details/76833835

本文针对一个特定的数学问题进行了解析，即求解2008的n次幂的所有正整数因子之和M，并进一步求解2008的M次幂对K取余的问题。通过巧妙的数学转换和编程实现，给出了具体的解题思路和代码实现。

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http://acm.hdu.edu.cn/showproblem.php?pid=1852

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 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

Sample Output

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。

题解：http://blog.youkuaiyun.com/weixin_36571742/article/details/76832624。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。

代码：

#include<bits/stdc++.h>
#define debug cout<<"aaa"<<endl
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define MIN_INT (-2147483647-1)
#define MAX_INT 2147483647
#define MAX_LL 9223372036854775807i64
#define MIN_LL (-9223372036854775807i64-1)
using namespace std;

const int N = 100000 + 5;
LL n,k,a,b,ans;

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

int main(){
	while(~scanf("%lld%lld",&n,&k)){
		if(n==0&&k==0){
			break;
		}
		a=quick(2,3*n+1,250*k);
		b=quick(251,n+1,250*k);//mod取250*k，目的是先要保留250这个因子，放在取余后再除
		a--,b--;
		ans=(a*b)%(250*k);
		ans/=250;
		ans=((ans%k)+k)%k;
		printf("%d\n",quick(2008,ans,k));
	}
	return 0;
}