poj 3233 Matrix Power Series - 矩阵快速幂

Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.

Input

　　The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

　　Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

　　这道题直接暴力呢。。$O\left(n^{3}k\right)$的时间复杂度，必死无疑。所以只能另寻出路，注意到矩阵满足分配律

矩阵乘法的右分配律 若$A$和$B$皆为$n$行$p$列的矩阵，$C$为$p$行$m$列的矩阵，则有

$(A + B)C = AB + AC$

　　左分配律差不多（因为矩阵乘法不满足交换律，所以矩阵乘法的分配律分为左分配律和右分配律）

　　证明就水水的写一下

　　对于$A + B$，我们有$\left(A + B \right )_{ij} = A_{ij} + B_{ij}$

　　那么

$\left[\left(A + B \right )C \right ]_{ij}\\=\sum_{i = 1}^{p}\left(A + B \right )_{ip}C_{pj}\\=\sum_{i = 1}^{p}A_{ip}C_{pj} + \sum_{i = 1}^{p}B_{ip}C_{pj}\\=\left(AC + BC \right )_{ij}$

　　所以$(A + B)C = AB + AC$

　　因为有了分配律这个神奇东西，现在计算$A + A^{2} + \cdots + A^{6}$的和，就等价于$\left(A + A^{2} + A^{3} \right ) + \left(A + A^{2} + A^{3} \right )A^{3}$

　　现在设$f\left(n \right ) = \sum_{i = 1}^{n}A^{i}$

　　那么有

$f\left ( n \right )=\left\{\begin{matrix}A\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(n = 1 \right )\\f\left(n - 1 \right ) + A^{n}\ \ \ \left(2 \nmid n \right ) \\ f\left(\frac{n}{2}\right) \left(I + A^{\frac{n}{2}} \right )\ \ \left(2 \mid n \right )  \end{matrix}\right.$

　　时间复杂度是$O\left(n^{3}\log^{2}k \right )$，但是如果代码写得比较丑就有TLE的风险，可以考虑一下以下优化

  1. 不要没事就作无用的初始化，很耗时间]
  2. 降低取模次数
  3. 适当地使用取模优化
  4. 减少无用的内存拷贝的次数

Code

  1 /**
  2  * poj
  3  * Problem#3233
  4  * Accepted
  5  * Time:2266ms
  6  * Memory:3892k 
  7  */
  8 #include<iostream>
  9 #include<cstdio>
 10 #include<cctype>
 11 #include<cstring>
 12 #include<cstdlib>
 13 #include<fstream>
 14 #include<sstream>
 15 #include<algorithm>
 16 #include<map>
 17 #include<set>
 18 #include<queue>
 19 #include<vector>
 20 #include<stack>
 21 using namespace std;
 22 typedef bool boolean;
 23 #define INF 0xfffffff
 24 #define smin(a, b) a = min(a, b)
 25 #define smax(a, b) a = max(a, b)
 26 template<typename T>
 27 inline void readInteger(T& u){
 28     char x;
 29     int aFlag = 1;
 30     while(!isdigit((x = getchar())) && x != '-');
 31     if(x == '-'){
 32         x = getchar();
 33         aFlag = -1;
 34     }
 35     for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
 36     ungetc(x, stdin);
 37     u *= aFlag;
 38 }
 39 
 40 typedef class Matrix {
 41     public:
 42         int* p;
 43         int lines, cols;
 44         int moder;
 45         Matrix():p(NULL), lines(0), cols(0), moder(1)    {    } 
 46         Matrix(int lines, int cols, int moder):lines(lines), cols(cols), moder(moder) {
 47             p = new int[(lines * cols)];
 48         }
 49         
 50         int* operator [](int pos) {
 51             return p + pos * cols;
 52         }
 53         
 54         Matrix operator *(Matrix b) {
 55             Matrix res(lines, b.cols, moder);
 56             for(int i = 0; i < lines; i++) {
 57                 for(int j = 0; j < b.cols; j++) {
 58                     res[i][j] = 0;
 59                     for(int k = 0; k < cols; k++) {
 60                         (res[i][j] += (*this)[i][k] * b[k][j] % moder) %= moder;
 61                     }
 62                 }
 63             }
 64             return res;
 65         }
 66         
 67         Matrix operator +(Matrix b) {
 68             Matrix res(lines, cols, moder);
 69             for(int i = 0; i < lines; i++)
 70                 for(int j = 0; j < cols; j++)
 71                     res[i][j] = ((*this)[i][j] + b[i][j]) % moder;
 72             return res;
 73         }
 74 }Matrix;
 75 
 76 template<typename T>
 77 T pow(T a, int pos) {
 78     if(pos == 1)    return a;
 79     T temp = pow(a, pos / 2);
 80     if(pos & 1)    return temp * temp * a;
 81     return temp * temp;
 82 }
 83 
 84 int n, k, m;
 85 Matrix a;
 86 
 87 inline void init() {
 88     readInteger(n);
 89     readInteger(k);
 90     readInteger(m);
 91     a = Matrix(n, n, m);
 92     for(int i = 0; i < n; i++) {
 93         for(int j = 0; j < n; j++) {
 94             readInteger(a[i][j]);
 95             a[i][j] %= m;
 96         }
 97     }
 98 }
 99 
100 template<typename T>
101 T pow_sum(T a, int pos) {
102     if(pos == 1)    return a;
103     T temp = pow_sum(a, pos / 2);
104     temp = temp + temp * pow(a, pos / 2);
105     if(pos & 1)    return temp + pow(a, pos);
106     return temp;
107 }
108 
109 Matrix res;
110 inline void solve() {
111     res = pow_sum(a, k);
112     for(int i = 0; i < n; i++) {
113         for(int j = 0; j < n; j++) {
114             printf("%d ", res[i][j]);
115         }
116         putchar('\n');
117     }
118 }
119 
120 int main() {
121     init();
122     solve();
123     return 0;
124 }

 

posted @ 2017-03-18 21:27  阿波罗2003  阅读(355)  评论(0)  编辑  收藏  举报