HDU 5451 Best Solver（数学）

思路：显然无理数是不能取模的，考虑An=(5+2sqrt(6))^n+(5-2sqrt(6))^n显然是一个整数，那么因为0<5-2sqrt(6)<1，而它们两者相加又是整数，那么就说明5+2sqrt(6)和5-2sqrt(6)的小数点部分是互补的，那么(5+2sqrt(6))^n+(5-2sqrt(6))^n-1其实就是5+2sqrt(6)的整数部分，就是结果了，然后就是求An的递推式了

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 48000;
int mod,a[maxn];
LL qpow(LL a,LL b,LL m) // 返回a^b % m
{
	LL ans = 1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%m;
		a = (a*a)%m;
		b>>=1;
	}
	return ans;
}

int gao(int mod)
{
	a[0]=10%mod;
	a[1]=98%mod;
	for(int i = 2;i<=maxn;i++)
	{
		a[i]=(a[i-1]*10-a[i-2]+mod)%mod;
		if(a[i-1]==a[0]&&a[i]==a[1])
			return i-1;
	}
}
int main()
{
   int T,cas=1;
   scanf("%d",&T);
   while(T--)
   {
	  LL n;
      scanf("%lld%d",&n,&mod);
      int x = gao(mod);
	  int res = qpow(2,n,x);
	  printf("Case #%d: %d\n",cas++,(a[res]-1+mod)%mod);
   }
}

Description

The so-called best problem solver can easily solve this problem, with his/her childhood sweetheart. 

It is known that 

. 

For a given integer 

 and a given prime number 

, print 

. (

 means the integer part of 

) 

Input

, indicating there are 

 test cases. 

Following are 

 lines, each containing two integers 

 and 

, as introduced above.

Output

 lines. 

Each line contains an integer representing 

.

Sample Input

70 463371 463373 46337 1 46337 21 46337 321 46337 4321 46337

Sample Output

Case #1: 97Case #2: 969Case #3: 16537Case #4: 969 Case #5: 40453 Case #6: 10211 Case #7: 17947