POJ3233Matrix Power Series(十大矩阵问题之三 + 二分+矩阵快速幂)
http://poj.org/problem?id=3233
Matrix Power Series
Time Limit: 3000MS | Memory Limit: 131072K | |
Total Submissions: 18658 | Accepted: 7895 |
Description
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
Source
POJ Monthly--2007.06.03, Huang, Jinsong
当n为奇数;假设 n = 7: A + A ^ 2 + A ^ 3 + A ^ 4 + A ^ 5 + A ^ 6 + A^7 = A + A ^ 2 + A ^ 3 + A ^ 3 * (A + A ^ 2 + A ^ 3) + A ^ 7
当n为偶数的时候就简单了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 6 using namespace std; 7 int n,m,k; 8 struct Mat 9 { 10 int mat[35][35]; 11 }; 12 Mat operator* (Mat x, Mat y) 13 { 14 Mat c; 15 memset(c.mat, 0, sizeof(c.mat)); 16 for(int t = 1; t <= n; t++) 17 { 18 for(int i = 1; i <= n; i++) 19 { 20 for(int j = 1; j <= n; j++) 21 c.mat[i][j] = (c.mat[i][j] + x.mat[i][t] % m * (y.mat[t][j] % m) ) % m; 22 } 23 } 24 return c; 25 } 26 Mat operator^ (Mat x, int y) 27 { 28 Mat c; 29 for(int i = 1; i <= n; i++) 30 for(int j = 1; j <= n; j++) 31 c.mat[i][j] = (i == j); 32 while(y) 33 { 34 if(y & 1) 35 c = c * x; 36 x = x * x; 37 y >>= 1; 38 } 39 return c; 40 } 41 Mat operator + (Mat x, Mat y) 42 { 43 for(int i = 1; i <= n; i++) 44 { 45 for(int j = 1; j <= n; j++) 46 x.mat[i][j] = ( x.mat[i][j] % m + y.mat[i][j] % m ) % m; 47 } 48 return x; 49 } 50 Mat dfs(int t,Mat temp) 51 { 52 if(t == 1) 53 return temp; 54 int mid = t / 2; 55 Mat c = dfs(mid, temp); 56 if(t & 1) 57 { 58 c = c + (temp ^ mid ) * c; 59 return c + (temp ^ t); //第一次交没加括号,查了好长时间的错,惭愧惭愧,其实codeblock都waring了,弱 60 } 61 else 62 return c + (temp ^ mid) * c; 63 } 64 int main() 65 { 66 67 scanf("%d%d%d", &n,&k,&m); 68 Mat a,c; 69 for(int i = 1; i <= n; i++) 70 for(int j = 1; j <= n; j++) 71 scanf("%d", &a.mat[i][j]); 72 c = dfs(k,a); 73 for(int i = 1; i <= n; i++) 74 { 75 for(int j = 1; j < n; j++) 76 printf("%d ", c.mat[i][j]); 77 printf("%d\n", c.mat[i][n]); 78 } 79 return 0; 80 }