POJ 3233 Matrix Power Series 经典矩阵快速幂＋二分

Posted 2021-01-18 ymzjj

任意门：http://poj.org/problem?id=3233

Matrix Power Series

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 + … + A^k.

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

 

题意概括：

给一个 N 维方阵 A ，求 A+A^2+A^3+ ... +A^k ，结果模 m；

解题思路：

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
", ans.m[i][N-1]);
80     }
81     return 0;
82 }

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