标签：display div tar ati cas \n each accept size
任意门：http://poj.org/problem?id=3233
| Time Limit: 3000MS | Memory Limit: 131072K | |
| Total Submissions: 28619 | Accepted: 11646 |
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
给一个 N 维方阵 A ，求 A+A^2+A^3+ ... +A^k ，结果模 m；
矩阵快速幂解决矩阵幂运算（本质是二分优化）；
求前缀和二分：
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)
应用这个式子后，规模k减小了一半。我们二分求出A^3后再递归地计算A + A^2 + A^3，即可得到原问题的答案。
AC code：
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 #define LL long long 7 using namespace std; 8 const int MAXN = 55; 9 int N, Mod; 10 11 struct mat 12 { 13 int m[MAXN][MAXN]; 14 }base; 15 16 mat mult(mat a, mat b) 17 { 18 mat res; 19 memset(res.m, 0, sizeof(res)); 20 for(int i = 0; i < N; i++) 21 for(int j = 0; j < N; j++){ 22 if(a.m[i][j]) 23 for(int k = 0; k < N; k++) 24 res.m[i][k] = (res.m[i][k] + a.m[i][j] * b.m[j][k])%Mod; 25 } 26 return res; 27 } 28 29 mat add(mat a, mat b) 30 { 31 mat res; 32 for(int i = 0; i < N; i++) 33 for(int j = 0; j < N; j++) 34 res.m[i][j] = (a.m[i][j] + b.m[i][j])%Mod; 35 return res; 36 } 37 38 mat qpow(mat a, int k) 39 { 40 mat ans; 41 memset(ans.m, 0, sizeof(ans.m)); 42 for(int i = 0; i < N; i++) ans.m[i][i] = 1; 43 44 while(k){ 45 if(k&1) ans = mult(ans, a); 46 k>>=1; 47 a=mult(a, a); 48 } 49 return ans; 50 } 51 52 mat solve(int K) 53 { 54 if(K == 1) return base; 55 mat res; 56 memset(res.m, 0, sizeof(res.m)); 57 for(int i = 0; i < N; i++) res.m[i][i] = 1; 58 59 res = add(res, qpow(base, K>>1)); 60 res = mult(res, solve(K>>1)); 61 if(K&1) res = add(res, qpow(base, K)); 62 63 return res; 64 } 65 66 int main() 67 { 68 int K; 69 mat ans; 70 scanf("%d %d %d", &N, &K, &Mod); 71 for(int i = 0; i < N; i++) 72 for(int j = 0; j < N; j++){ 73 scanf("%d", &base.m[i][j]); 74 } 75 ans = solve(K); 76 for(int i = 0; i < N; i++){ 77 for(int j = 0; j < N-1; j++) 78 printf("%d ", ans.m[i][j]); 79 printf("%d\n", ans.m[i][N-1]); 80 } 81 return 0; 82 }
POJ 3233 Matrix Power Series 【经典矩阵快速幂＋二分】
标签：display div tar ati cas \n each accept size
原文地址：https://www.cnblogs.com/ymzjj/p/9919832.html
