hdu 2276 矩阵快速幂&&循环矩阵优化

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我们可以用+1之后对2取余来模拟灯的变化。

然后首先构造矩阵：

1 0 0 0 0 0 1
0 1 1 0 0 0 0
0 0 1 1 0 0 0
0 0 0 1 1 0 0
0 0 0 0 1 1 0
0 0 0 0 0 1 1

可以推广到大小为N的矩阵

然后这是一个循环矩阵，要优化一下乘法

AC代码如下：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int MAX_N = 110;;int N, M;void multipy( int a[][MAX_N], int am, int an, int b[][MAX_N], int bm, int bn, int c[][MAX_N] ){    for( int i = 1; i <= am; i++ ){        for( int j = 1; j <= bn; j++ ){            c[i][j] = 0;            for( int k = 1; k <= an; k++ ){                c[i][j] = ( c[i][j] + a[i][k] * b[k][j] ) % 2;            }        }    }}void multipy1( int a[][MAX_N], int am, int an, int b[][MAX_N], int bm, int bn, int c[][MAX_N] ){    for( int i = 1; i <= 1; i++ ){        for( int j = 1; j <= bn; j++ ){            c[i][j] = 0;            for( int k = 1; k <= an; k++ ){                c[i][j] = ( c[i][j] + a[i][k] * b[k][j] ) % 2;            }        }    }    for( int i = 2; i <= am; i++ ){        c[i][1] = c[i-1][bn];        for( int j = 2; j <= bn; j++ ){            c[i][j] = c[i-1][j-1];        }    }}void get_matrix_pow( int a[][MAX_N], int n ){    int ans[MAX_N][MAX_N] = {0};    int temp[MAX_N][MAX_N];    for( int i = 1; i <= N; i++ ){        ans[i][i] = 1;    }    while( n ){        if( n % 2 ){            multipy1( ans, N, N, a, N, N, temp );            memcpy( ans, temp, sizeof( int ) * MAX_N * MAX_N );        }        multipy1( a, N, N, a, N, N, temp );        memcpy( a, temp, sizeof( int ) * MAX_N * MAX_N );        n /= 2;    }    memcpy( a, ans, sizeof( int ) * MAX_N * MAX_N );}int main(){    char s[MAX_N];    int a[MAX_N][MAX_N];    int ans[MAX_N][MAX_N];    int temp[MAX_N][MAX_N];    while( scanf( "%d", &M ) != EOF ){        scanf( "%s", s );        N = strlen( s );        memset( a, 0, sizeof( a ) );        a[1][1] = a[1][N] = 1;        for( int i = 2; i <= N; i++ ){            a[i][i] = a[i][i-1] = 1;        }/*        cout << "--------------" << endl;        for( int i = 1; i <= N; i++ ){            for( int j = 1; j <= N; j++ ){                cout << a[i][j];            }            cout << endl;        }        cout << "--------------" << endl;*/        get_matrix_pow( a, M );/*        cout << "--------------" << endl;        for( int i = 1; i <= N; i++ ){            for( int j = 1; j <= N; j++ ){                cout << a[i][j];            }            cout << endl;        }        cout << "--------------" << endl;*/        for( int i = 1; i <= N; i++ ){            ans[i][1] = s[i-1] - '0';        }        multipy( a, N, N, ans, N, 1, temp );        for( int i = 1; i <= N; i++ ){            printf( "%c", temp[i][1] + '0' );        }        puts("");    }    return 0;}

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