POJ 3233 Matrix Power Series

题目链接：http://poj.org/problem?id=3233

题目：给出一个 n×n 的矩阵 A 和一个整数k, 求 S = A + A² + A³ + … + A^k.  

其中n (n ≤ 30), k (k ≤ 10⁹) , m (m < 10⁴).

分析：两次二分，A^i可以用矩阵快速幂，然后求和二分。比如，当k=6时，有：A + A^2 + A^3 + A^4 + A^5 + A^6 =(A + A^2 + A^3) + A^3*(A + A^2 + A^3)

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我用的递归求和，程序很杯具的跑了2s……等有空再改改吧，写的真的很糟糕，主要是想练习一下二分。

#include <cstdio>
#include <cstring>

const int MAXN = 35;

struct Mat
{
    int array[MAXN][MAXN];
    friend Mat operator*( Mat &a, Mat &b );
    friend Mat operator+( Mat &a, Mat &b );
    friend Mat operator^( Mat a, int k);
};

int n, MOD;
Mat data;
Mat UNIT;

Mat operator*( Mat &a, Mat &b )
{
    Mat temp;

    memset( temp.array, 0, sizeof(temp.array) );

    for ( int i = 0; i < n; i++ )
       for ( int j = 0; j < n; j++ )
           for ( int k = 0; k < n; k++ )
              temp.array[i][j] = ( temp.array[i][j] + a.array[i][k] * b.array[k][j]%MOD )%MOD;

    return temp;
}

Mat operator+( Mat &a, Mat &b )
{
    Mat temp;

    memset( temp.array, 0, sizeof(temp.array) );

    for ( int i = 0; i < n; i++ )
       for ( int j = 0; j < n; j++ )
            temp.array[i][j] = ( a.array[i][j] + b.array[i][j] ) % MOD;

    return temp;
}

Mat operator^( Mat a, int k )
{
    Mat unit = UNIT;

    while ( k )
    {
        if ( k&1 ) unit = unit * a;
        a = a * a;
        k >>= 1;
    }

    return unit;
}

Mat Sum( int k )
{
    Mat temp;
    if ( k == 1 ) return temp = data;

    memset( temp.array, 0, sizeof(temp.array) );

    int tp = k >> 1;

    Mat temp2 = Sum( tp );
    Mat temp3 = data ^ tp;

    Mat temp1 = temp2 * temp3;
    temp = temp2 + temp1;

    if ( k&1 )
    {
        Mat temp4 = data ^ k;
        temp = temp + temp4;
    }

    return temp;
}

int main()
{
    int k;
    memset( UNIT.array, 0, sizeof(UNIT.array) );
    for ( int i = 0; i < MAXN; i++ )
        UNIT.array[i][i] = 1;

    while ( scanf("%d%d%d", &n, &k, &MOD) != EOF )
    {
        for ( int i = 0; i < n; i++ )
           for ( int j = 0; j < n; j++ )
            {
                scanf( "%d", &data.array[i][j] );
                data.array[i][j] %= MOD;
            }

        Mat ans = Sum(k);

        for ( int i = 0; i < n; i++ )
        {
            printf( "%d", ans.array[i][0] );
            for ( int j = 1; j < n; j++ )
               printf( " %d", ans.array[i][j] );
            putchar('\n');
        }

    }
    return 0;
}