HDU 4291 A Short problem(矩阵快速幂取模求循环节)_模取幂结果是个循环-优快云博客

该博客探讨了一道编程题目，涉及利用矩阵快速幂的方法解决计算g(g(g(n))) mod 10^9 + 7的问题，其中g(n)由递推关系3g(n - 1) + g(n - 2)给出，初始条件为g(1) = 1, g(0) = 0。博主注意到由于三层递归的存在，问题中可能存在循环节，因此重点在于如何找到这个循环节。" 76991160,5112153,自动化测试实践指南,"['自动化测试', '测试工具', '测试框架', '持续集成', '软件质量']

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A Short problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1554 Accepted Submission(s): 566

Problem Description

　　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given n (1 <= n <= 10¹⁸), You should solve for
g(g(g(n))) mod 10⁹ + 7
　　where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0

Input

　　There are several test cases. For each test case there is an integer n in a single line.
　　Please process until EOF (End Of File).

Output

　　For each test case, please print a single line with a integer, the corresponding answer to this case.

Sample Input

0
1
2

Sample Output

0
1
42837

一眼看就会想到矩阵快速幂求模，但是仔细一看数据和g(g(g(x)))有三层就想到了这一定有循环节。

求循环节

#include <stdio.h>
#define ll __int64
#define mod1 1000000007
#define mod2 222222224
#define mod3 183120
int main(){
	ll a=0,b=1,c,i;
	for(i=2;;i++){//求循环节 
		c=(b*3+a)%mod1;//mod
		a=b;
		b=c;
		if(a == 0 && b == 1){
			printf("%I64d\n",i-1);//mod2=222222224,mod3=183120
			break;
		}
	}
    return 0;
}

AC代码：

#include <stdio.h>
#include <string.h>
#define ll __int64
#define Max 2 //矩阵大小,注意能小就小  
ll n;
int mod[]={183120,222222224,1000000007};
struct Matrix{
      ll  mm[Max][Max];
}A,I,W;

Matrix matrixmul(Matrix a,Matrix b,int mod) //矩阵乘法
{
       int i,j,k;
       Matrix c;
       for (i = 0 ; i < 2; i++)
           for (j = 0; j <2;j++){
                 c.mm[i][j] = 0;
                 for (k = 0; k <2; k++)
                     c.mm[i][j] += (a.mm[i][k] * b.mm[k][j])%mod;
				  c.mm[i][j]%=mod;
             }
       return c;
}
ll quickpagow(ll n,int mod){
       Matrix m = A, b = I;//b是单位矩阵
       while (n >= 1){
             if (n & 1)
                b = matrixmul(b,m,mod);
             n = n >> 1;
             m = matrixmul(m,m,mod);
       }
	   b=matrixmul(W,b,mod);
       return b.mm[0][0]%mod;
}

int main (){
	ll i;

	A.mm[0][0]=0,A.mm[0][1]=1;
	A.mm[1][0]=1,A.mm[1][1]=3;

	I.mm[0][0]=1,I.mm[0][1]=0;
	I.mm[1][0]=0,I.mm[1][1]=1;

	W.mm[0][0]=3,W.mm[0][1]=10;
	W.mm[1][0]=0,W.mm[1][1]=0;
	while(~scanf("%I64d",&n)){
		for(i=0;i<3;i++){
			if(n>=2)
				n=quickpagow(n-2,mod[i]);
		}
		printf("%I64d\n",n);
	}
	return 0;
}