hdu5863 cjj's string game

矩阵快速幂

#include<bits/stdc++.h>

using namespace std;

const int INF = 0x3f3f3f3f;

const int MOD = 1e9+7;

int N,M,K;

struct Matrix{

	int a[25][25];

	int n;

	Matrix(int _n=0){

		memset(a,0,sizeof(a));

		n = _n;

	}

	Matrix operator *(const Matrix &T) const{

		Matrix ans = Matrix(n);

		for(int i = 0; i < n; ++i)

			for(int j = 0; j < n; ++j)

				for(int k = 0; k < n; ++k)

					ans.a[i][j] = (ans.a[i][j] + 1ll*a[i][k]*T.a[k][j]%MOD) % MOD;

		return ans;

	}

};

Matrix operator ^(Matrix x,int y){

	Matrix ans = Matrix(x.n);

	for(int i = 0; i < x.n; ++i) ans.a[i][i] = 1;

	while(y){

		if(y&1) ans=ans*x;

		y >>= 1; x = x*x;

	}

	return ans;

}

int main(){

	int T; scanf("%d",&T);

	while(T--) {

		scanf("%d %d %d",&N,&M,&K);

		Matrix A = Matrix(M*2+1);

		A.a[0][0] = 1;

		Matrix P = Matrix(M*2+1);

		for(int i = 0; i < M; ++i) P.a[i][0] = K*K-K;

		for(int i = 1; i < M; ++i)  P.a[i-1][i] = K;

		P.a[M-1][M*2] = K; P.a[M*2-1][M*2] = K;

		for(int i = M; i < M*2+1; ++i) P.a[i][M] = K*K-K;

		for(int i = M+1; i < M*2; ++i) P.a[i-1][i] = K;

		P = P^N;

		P = A*P;

		int ans = 0;

		for(int i = M; i < M*2+1; ++i) ans = (ans + P.a[0][i]) % MOD;

		printf("%d\n",ans);

	}

	return 0;

}