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。

  1 #include <cstdio>

  2 #include <cstring>

  3 

  4 const int MAXN = 35;

  5 

  6 struct Mat

  7 {

  8     int array[MAXN][MAXN];

  9     friend Mat operator*( Mat &a, Mat &b );

 10     friend Mat operator+( Mat &a, Mat &b );

 11     friend Mat operator^( Mat a, int k);

 12 };

 13 

 14 int n, MOD;

 15 Mat data;

 16 Mat UNIT;

 17 

 18 Mat operator*( Mat &a, Mat &b )

 19 {

 20     Mat temp;

 21 

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

 23 

 24     for ( int i = 0; i < n; i++ )

 25        for ( int j = 0; j < n; j++ )

 26            for ( int k = 0; k < n; k++ )

 27               temp.array[i][j] = ( temp.array[i][j] + a.array[i][k] * b.array[k][j]%MOD )%MOD;

 28 

 29     return temp;

 30 }

 31 

 32 Mat operator+( Mat &a, Mat &b )

 33 {

 34     Mat temp;

 35 

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

 37 

 38     for ( int i = 0; i < n; i++ )

 39        for ( int j = 0; j < n; j++ )

 40             temp.array[i][j] = ( a.array[i][j] + b.array[i][j] ) % MOD;

 41 

 42     return temp;

 43 }

 44 

 45 Mat operator^( Mat a, int k )

 46 {

 47     Mat unit = UNIT;

 48 

 49     while ( k )

 50     {

 51         if ( k&1 ) unit = unit * a;

 52         a = a * a;

 53         k >>= 1;

 54     }

 55 

 56     return unit;

 57 }

 58 

 59 Mat Sum( int k )

 60 {

 61     Mat temp;

 62     if ( k == 1 ) return temp = data;

 63 

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

 65 

 66     int tp = k >> 1;

 67 

 68     Mat temp2 = Sum( tp );

 69     Mat temp3 = data ^ tp;

 70 

 71     Mat temp1 = temp2 * temp3;

 72     temp = temp2 + temp1;

 73 

 74     if ( k&1 )

 75     {

 76         Mat temp4 = data ^ k;

 77         temp = temp + temp4;

 78     }

 79 

 80     return temp;

 81 }

 82 

 83 int main()

 84 {

 85     int k;

 86     memset( UNIT.array, 0, sizeof(UNIT.array) );

 87     for ( int i = 0; i < MAXN; i++ )

 88         UNIT.array[i][i] = 1;

 89 

 90     while ( scanf("%d%d%d", &n, &k, &MOD) != EOF )

 91     {

 92         for ( int i = 0; i < n; i++ )

 93            for ( int j = 0; j < n; j++ )

 94             {

 95                 scanf( "%d", &data.array[i][j] );

 96                 data.array[i][j] %= MOD;

 97             }

 98 

 99         Mat ans = Sum(k);

100 

101         for ( int i = 0; i < n; i++ )

102         {

103             printf( "%d", ans.array[i][0] );

104             for ( int j = 1; j < n; j++ )

105                printf( " %d", ans.array[i][j] );

106             putchar('\n');

107         }

108 

109     }

110     return 0;

111 }