快速幂应用

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本文探讨了矩阵快速幂算法及其在多项式乘法、GoogleCodeJam问题解决、斐波那契数列求解、Block涂色方案计算等领域的应用。通过矩阵快速幂算法优化复杂度，实现高效计算。文章详细介绍了算法原理、代码实现及实际案例分析。

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矩阵快速幂自乘

typedef vector< int > vec;
typedef vector< vector< int > > mat;


mat multi( const mat& A, const 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 quick_pow( mat& A, LL n ){

    int row = A.size();
    int col = A[0].size();
    mat B( row, vec( col ) );

    for( int i = 0; i < row; ++i ){
        B[i][i] = 1;
    }

    while( n > 0 ){

        if( n & 1 )
            B = multi( B, A );

        A = multi( A, A );
        n >>= 1;

    }

    return B;

}

Google Code Jam -- Number( 2008 Round C )

求 ( 3 + 5 ^ 0.5 ) ^ N ( 1 <= N <= 10 ^ 9 ) 整数部分的最后三位，不足三位数的话首部填零。

( 3 + 5 ^ 0.5 ) ^ N 都可以变形为 ( A + B * ( 5 ^ 0.5 ) ) 的结构

( 3 + 5 ^ 0.5 ) ^ ( N + 1 ) = ( 3 + 5 ^ 0.5 ) * ( 3 + 5 ^ 0.5 ) ^ N

= ( 3 + 5 ^ 0.5 ) * ( A + B * ( 5 ^ 0.5 ) )

= ( A' + B' * ( 5 ^ 0.5 ) )

A' = 3 * A + 5 * B

B' = A + 3 * B

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;

#define MOD 1000
#define LL long long

typedef vector< int > vec;
typedef vector< vector< int > > mat;


mat multi( const mat& A, const 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 quick_pow( mat& A, LL n ){

    int row = A.size();
    int col = A[0].size();
    mat B( row, vec( col ) );

    for( int i = 0; i < row; ++i ){
        B[i][i] = 1;
    }

    while( n > 0 ){

        if( n & 1 )
            B = multi( B, A );

        A = multi( A, A );
        n >>= 1;

    }

    return B;

}


int main(){

    LL n;
    mat A( 2, vec( 2, 0 ) );

    A[0][0] = 3; A[0][1] = 5;
    A[1][0] = 1; A[1][1] = 3;

    cin >> n;

    A = quick_pow( A, n );

    printf( "%03d\n", ( A[0][0] * 2 + MOD - 1 ) % MOD );

    return 0;
}

POJ 3070 斐波那契

求解第 N( 1 <= N <= 10 ^ 9 ) 个斐波那契数

| F( n + 2 ) | | 1 1 | | F( n + 1 ) |

| | = | | * | |

| F( n + 1 ) | | 1 0 | | F( n ) |

| F( n + 1 ) | | F( 1 ) |

| | = A ^ n * | |

| F( n ) | | F( 0 ) |

#include <iostream>
#include <vector>
using namespace std;

#define MOD 10007
#define LL long long

typedef vector< int > vec;
typedef vector< vector< int > > mat;


mat multi( const mat& A, const 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 quick_pow( mat& A, LL n ){

    int row = A.size();
    int col = A[0].size();
    mat B( row, vec( col ) );

    for( int i = 0; i < row; ++i ){
        B[i][i] = 1;
    }

    while( n > 0 ){

        if( n & 1 )
            B = multi( B, A );

        A = multi( A, A );
        n >>= 1;

    }

    return B;

}


int main(){

    int test;

    cin >> test;

    while( test-- ){

        LL n;
        mat A( 3, vec ( 3 ) );

        A[0][0] = 2; A[0][1] = 1; A[0][2] = 0;
        A[1][0] = 2; A[1][1] = 2; A[1][2] = 2;
        A[2][0] = 0; A[2][1] = 1; A[2][2] = 2;

        cin >> n;

        A = quick_pow( A, n );

        cout << A[0][0] << endl;

    }
    return 0;
}

POJ 3734 Blocks

一行 N 个方块，每块只能用用 A, B, C, D 四种之一颜色涂满，求含有偶数个颜色 A 和偶数个颜色 B 的方块的涂色方案的个数。

递推：

涂到第 i 块的时后，A, B 都是偶数的方案个数为 X，一奇一偶方案个数为 Y，全为奇数方案个数为 Z

那么低 i + 1 的可能性：

X' = 2 * X + Y

Y' = 2 * X + 2 * Y + 2 * Z

Z' = Y + 2 * Z

#include <iostream>
#include <vector>
using namespace std;

#define MOD 10007
#define LL long long

typedef vector< int > vec;
typedef vector< vector< int > > mat;


mat multi( const mat& A, const 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 quick_pow( mat& A, LL n ){

    int row = A.size();
    int col = A[0].size();
    mat B( row, vec( col ) );

    for( int i = 0; i < row; ++i ){
        B[i][i] = 1;
    }

    while( n > 0 ){

        if( n & 1 )
            B = multi( B, A );

        A = multi( A, A );
        n >>= 1;

    }

    return B;

}


int main(){

    int test;

    cin >> test;

    while( test-- ){

        LL n;
        mat A( 3, vec ( 3 ) );

        A[0][0] = 2; A[0][1] = 1; A[0][2] = 0;
        A[1][0] = 2; A[1][1] = 2; A[1][2] = 2;
        A[2][0] = 0; A[2][1] = 1; A[2][2] = 2;

        cin >> n;

        A = quick_pow( A, n );

        cout << A[0][0] << endl;

    }
    return 0;
}