poj 3233 矩阵快速幂

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十个利用矩阵乘法解决的经典题目的最后一题

终于刷完十道题了，基本都是一个思路：二分优化！

a^1 + a^2 + a^3 + a^4 + a^5 + a^6 + a^7 + a^8 用二分变成 a^1 + a^2 + a^3 + a^4 + a^4 *（ a^1 + a^2 + a^3 + a^4 ）

a^1 + a^2 + a^3 + a^4 + a^5 + a^6 + a^7 + a^8 + a^9 用二分变成 a^1 + a^2 + a^3 + a^4 + a^4 *（ a^1 + a^2 + a^3 + a^4 ）+ a^9

AC代码如下：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int MAX_N = 33;int MOD, N, K;void multipy( int a[MAX_N][MAX_N], int am, int an, int b[MAX_N][MAX_N], int bm, int bn, int c[MAX_N][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] ) % MOD;            }        }    }}void add_matrix( int a[MAX_N][MAX_N], int b[MAX_N][MAX_N], int m, int n ){    for( int i = 1; i <= m; i++ ){        for( int j = 1; j <= n; j++ ){            a[i][j] = ( a[i][j] + b[i][j] ) % MOD;        }    }}void get_matrix_pow( int a[MAX_N][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 & 1 ){            multipy( ans, N, N, a, N, N, temp );            memcpy( ans, temp, sizeof( int ) * MAX_N * MAX_N );        }        multipy( 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 );}void solve( int a[MAX_N][MAX_N], int b[MAX_N][MAX_N], int n ){    if( n == 1 ){        memcpy( a, b, sizeof( int ) * MAX_N * MAX_N );        return;    }    if( n == 0 ){        for( int i = 1; i <= N; i++ ){            for( int j = 1; j <= N; j++ ){                a[i][j] = 0;            }        }        for( int i = 1; i <= N; i++ ){            a[i][i] = 1;        }        return;    }    int temp1[MAX_N][MAX_N], temp2[MAX_N][MAX_N];    solve( a, b, n / 2 );    memcpy( temp1, b, sizeof( int ) * MAX_N * MAX_N );    get_matrix_pow( temp1, n / 2 );    multipy( temp1, N, N, a, N, N, temp2 );    add_matrix( a, temp2, N, N );    if( n & 1 ){        memcpy( temp1, b, sizeof( int ) * MAX_N * MAX_N );        get_matrix_pow( temp1, n );        add_matrix( a, temp1, N, N );    }}int main(){    int a[MAX_N][MAX_N], b[MAX_N][MAX_N];    while( scanf( "%d%d%d", &N, &K, &MOD ) != EOF ){        for( int i = 1; i <= N; i++ ){            for( int j = 1; j <= N; j++ ){                scanf( "%d", &b[i][j] );            }        }        solve( a, b, K );        for( int i = 1; i <= N; i++ ){            printf( "%d", a[i][1] );            for( int j = 2; j <= N; j++ ){                printf( " %d", a[i][j] );            }            printf( "\n" );        }    }    return 0;}

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