poj 3233 矩阵快速幂
地址 http://poj.org/problem?id=3233
大意是n维数组 最多k次方 结果模m的相加和是多少
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
Sample Input
2 2 4 0 1 1 1
Sample Output
1 2 2 3
题解
矩阵逐步的相乘然后相加是不可以 但是矩阵也有类似快速幂的做法
/*
A + A^2 =A(I+A)
A + A^2 + A^3 + A^4 = (A + A^2)(I + A^2)
记做sum(n) = A +A^2 +A^3 +...+A^n
如果n是偶数 sum(n) = sum(n/2)(I+A^(n/2))
如果n是奇数 sum(n) = sum(n-1) + A^n
= sum((n-1)/2)(I+A^((n-1)/2)) + A^n
*/
代码如下
1 #include <iostream> 2 #include <cstring> 3 4 using namespace std; 5 6 7 struct matrix { 8 int data[35][35]; 9 }; 10 11 int n = 0; 12 int m = 0; 13 int k = 0; 14 15 //矩阵乘法 16 matrix mul(matrix a, matrix b) 17 { 18 matrix c; 19 memset(c.data, 0, sizeof(c.data)); 20 for (int i = 1; i <= n; i++) { 21 for (int j = 1; j <= n; j++) { 22 for (int k = 1; k <= n; k++) { 23 c.data[i][j] = (c.data[i][j] + 1ll * a.data[i][k] * b.data[k][j]) % m; 24 } 25 } 26 } 27 28 return c; 29 } 30 31 //矩阵加法 32 matrix add(matrix a, matrix b) { 33 for (int i = 1; i <= n; i++) { 34 for (int j = 1; j <= n; j++) { 35 a.data[i][j] = (a.data[i][j] + b.data[i][j])%m; 36 } 37 } 38 return a; 39 } 40 41 //矩阵快速幂 42 matrix quickpow(matrix a, int k) { 43 matrix c; 44 memset(c.data, 0, sizeof(c.data)); 45 for (int i = 1; i <= n; i++) 46 c.data[i][i] = 1; 47 while (k) { 48 if (k & 1) c = mul(c, a); 49 k >>= 1; 50 a = mul(a, a); 51 } 52 return c; 53 } 54 55 //正式计算 sum k 56 matrix sum(matrix a, int k) { 57 if (k == 1) return a; 58 matrix c; 59 memset(c.data, 0, sizeof(c.data)); 60 for (int i = 1; i <= n; i++) 61 c.data[i][i] = 1; 62 c = add(c, quickpow(a, k >> 1)); 63 c = mul(c, sum(a, k >> 1)); 64 if (k & 1) c = add(c, quickpow(a, k)); 65 return c; 66 } 67 68 69 70 71 /* 72 A + A^2 =A(I+A) 73 74 A + A^2 + A^3 + A^4 = (A + A^2)(I + A^2) 75 记做sum(n) = A +A^2 +A^3 +...+A^n 76 如果n是偶数 sum(n) = sum(n/2)(I+A^(n/2)) 77 如果n是奇数 sum(n) = sum(n-1) + A^n 78 = sum((n-1)/2)(I+A^((n-1)/2)) + A^n 79 */ 80 81 int main() 82 { 83 matrix mat; 84 cin >> n; 85 cin >> k; 86 cin >> m; 87 88 for (int i = 1; i <= n; i++) { 89 for (int j = 1; j <= n; j++) { 90 cin >> mat.data[i][j]; 91 } 92 } 93 94 matrix ret = sum(mat, k); 95 96 97 98 for (int i = 1; i <= n; i++) { 99 for (int j = 1; j <= n; j++) { 100 cout << ret.data[i][j] << " "; 101 } 102 cout << endl; 103 } 104 }
作 者: itdef
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欢迎转帖 请保持文本完整并注明出处
技术博客 http://www.cnblogs.com/itdef/
B站算法视频题解
https://space.bilibili.com/18508846
qq 151435887
gitee https://gitee.com/def/
欢迎c c++ 算法爱好者 windows驱动爱好者 服务器程序员沟通交流
如果觉得不错,欢迎点赞,你的鼓励就是我的动力