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poj2663 矩阵快速幂

2024-07-23 11:00:50   221人阅读

Tri Tiling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7841 Accepted: 4113

Description

In how many ways can you tile a 3xn rectangle with 2x1 dominoes? 
Here is a sample tiling of a 3x12 rectangle. 

Input

Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 <= n <= 30.

Output

For each test case, output one integer number giving the number of possible tilings.

Sample Input

2812-1

Sample Output

31532131

思路:只有八种转移状态,设初始状态为0(全空),得到状态7(111全染)
#include <cstdio>#include <cstring>#include <vector>using namespace std;typedef vector<long long > vec;typedef vector<vec> mat;const int mod=(1<<31-1);mat mul(mat &A,mat &B){    mat C(A.size(),vec(B[0].size()));    for(int i=0;i<A.size();i++){        for(int k=0;k<B.size();k++){            for(int j=0;j<B[0].size();j++){                C[i][j]=(C[i][j]+A[i][k]*B[k][j])%mod;            }        }    }    return C;}mat pow(mat A,long long n){    mat B(A.size(),vec(A.size()));    for(int i=0;i<A.size();i++){        B[i][i]=1;    }    while(n>0){        if(n&1)B=mul(B,A);        A=mul(A,A);        n>>=1;    }    return B;}void calc(int m){    mat m1(8,vec(8,0)),E(8,vec(8,0));    E[0][7]=1;    E[1][6]=1;    E[2][5]=1;    E[3][4]=1;    E[4][3]=1;E[4][7]=1;    E[5][2]=1;    E[6][1]=E[6][7]=1;    E[7][0]=1;E[7][4]=E[7][6]=1;    E=pow(E,m);    /*for(int i=0;i<E[0].size();i++){        for(int j=0;j<E.size();j++){            printf("%d ",E[i][j]);        }        puts("");    }*/    printf("%I64d\n",E[0][7]);}int main(){    int m=0;    while(scanf("%d",&m)==1&&m!=-1){        calc(m+1);    }    return 0;}

　　

poj2663 矩阵快速幂


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