• A Simple Math Problem (矩阵快速幂)


    Lele now is thinking about a simple function f(x). 

    If x < 10 f(x) = x. 
    If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); 
    And ai(0<=i<=9) can only be 0 or 1 . 

    Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m. 

    Input

    The problem contains mutiple test cases.Please process to the end of file. 
    In each case, there will be two lines. 
    In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
    In the second line , there are ten integers represent a0 ~ a9. 

    Output

    For each case, output f(k) % m in one line.

    Sample Input

    10 9999
    1 1 1 1 1 1 1 1 1 1
    20 500
    1 0 1 0 1 0 1 0 1 0

    Sample Output

    45
    104

    代码:递推式套矩阵快速幂即可,注意多组输入

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    
    using namespace std;
    
    const int N=10;
    
    int n,mod;
    int temp[N][N];
    int res[N][N],a[N][N];
    void mul(int a[][N],int b[][N]) {
    	memset(temp,0,sizeof(temp));
    	for(int i=0; i<N; i++)
    		for(int j=0; j<N; j++)
    			for(int k=0; k<N; k++)
    				temp[i][j]=(temp[i][j]+a[i][k]*b[k][j]%mod)%mod;
    	for(int i=0; i<N; i++)
    		for(int j=0; j<N; j++)
    			a[i][j]=temp[i][j];
    	return ;
    }
    void QuickPow(int nn) {
    	memset(res,0,sizeof(res));
    	for(int i=0; i<N; i++)
    		res[i][i]=1;
    	while(nn) {
    		if(nn&1)
    			mul(res,a);
    		mul(a,a);
    		nn>>=1;
    	}
    	return ;
    }
    int main() {
    	while(~scanf("%d %d",&n,&mod)) {
    		memset(a,0,sizeof(a));
    		for(int i=0; i<N; i++)
    			scanf("%d",&a[0][i]);
    		for(int i=1; i<N; i++)
    			a[i][i-1]=1;
    		if(n<10) printf("%d
    ",n%mod);
    		else {
    			QuickPow(n-9);
    			int ans=0;
    			for(int i=0; i<N; i++)
    				ans+=res[0][i]*(9-i)%mod;
    			printf("%d
    ",ans%mod);
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781891.html
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