HDU 1575 Tr A 【矩阵经典2 矩阵快速幂入门】

hdoj 1575 Tr A (矩阵快速幂) Tr A Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4549    Accepted Submission(s): 3428 Problem Description A为一个方阵,则Tr A表示A的迹(就是主 阅读详情

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=1575

Tr A

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7572    Accepted Submission(s): 5539


Problem Description
A为一个方阵,则Tr A表示A的迹(就是主对角线上各项的和),现要求Tr(A^k)%9973。
 

 

Input
数据的第一行是一个T,表示有T组数据。
每组数据的第一行有n(2 <= n <= 10)和k(2 <= k < 10^9)两个数据。接下来有n行,每行有n个数据,每个数据的范围是[0,9],表示方阵A的内容。
 

 

Output
对应每组数据,输出Tr(A^k)%9973。
 

 

Sample Input
2
2 2
1 0
0 1
3 99999999
1 2 3
4 5 6
7 8 9
 

 

Sample Output
2
2686
 

 题意概括:如题干

解题思路:矩阵快速幂(本质是二分优化快速进行幂运算)矩阵快速幂模板题

注意:单位矩阵初始化

 

AC code:

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cmath>
 5 #define LL long long
 6 using namespace std;
 7 const int MAXN = 11;
 8 const int Mod = 9973;
 9 int N;
10 struct mat
11 {
12     int m[MAXN][MAXN];
13 }base;
14 
15 mat muti(mat a, mat b)
16 {
17     mat res;
18     memset(res.m, 0, sizeof(res.m));
19     for(int i = 1; i <= N; i++)
20     for(int j = 1; j <= N; j++){
21             if(a.m[i][j]){
22                 for(int k = 1; k <= N; k++){
23                     res.m[i][k] = (res.m[i][k] + a.m[i][j]*b.m[j][k])%Mod;
24                 }
25             }
26         }
27 
28     return res;
29 }
30 
31 mat qpow(mat a, int n)
32 {
33     mat res;
34     memset(res.m, 0, sizeof(res.m));
35     for(int i = 1; i <= N; i++) res.m[i][i] = 1;
36     while(n){
37         if(n&1) res = muti(res, a);
38         n>>=1;
39         a = muti(a, a);
40     }
41     return res;
42 }
43 
44 
45 int main()
46 {
47     int K, T_case;
48     scanf("%d", &T_case);
49     while(T_case--){
50         memset(base.m, 0, sizeof(base.m));
51         scanf("%d %d", &N, &K);
52         for(int i = 1; i <= N; i++){
53             for(int j = 1; j <= N; j++){
54                 scanf("%d", &base.m[i][j]);
55             }
56         }
57         base = qpow(base, K);
58         int ans = 0;
59         for(int i = 1; i <= N; i++){
60             ans = (ans + base.m[i][i])%Mod;
61         }
62         printf("%d\n", ans);
63     }
64 
65     return 0;
66 }
View Code

 

矩阵快速幂详解 设想这样一个场景,面试官给了你一道算法题,你很快确定这是一道递推问题,并给出了 O(n) 的解题方法,然而面试官却继续问:“还能继续优化吗?” 这样类似的场景并不少见,因为算法不仅追求「正确」,还追求「效率」,而这也正是优化方法的意义。 本文即将介绍的「矩阵快速幂」便是一种常见的优化递推的方法,能将 O(n) 的递推过程加速至 O(log(n)),使效率显著提升。 阅读详情

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题目地址:HDU 1575 矩阵高速幂裸题。 初学矩阵高速幂。曾经学过高速幂。今天一看矩阵高速幂,原来其原理是一样的,这就好办多了。都是利用二分的思想不断的乘。仅仅只是把数字变成了矩阵而已。 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include ...

aofan9566的博客 117

HDU1575 Tr A(矩阵快速幂)

A - Tr A Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description A为一个方阵,则Tr A表示A的迹(就是主对角线上各项的和),现要求Tr(A^k)%9973。    Input

叶孤心的专栏 555
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