poj 3233 矩阵快速幂

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 + … + A^k.

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 }