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 + … + 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

 #include <iostream>
 #include <cstring>
 
 using namespace std;
 
 struct matrix {
     int data[][];
 };
 
 int n = ;
 int m = ;
 int k = ;
 
 //矩阵乘法
 matrix mul(matrix a, matrix b)
 {
     matrix c;
     memset(c.data, , sizeof(c.data));
     for (int i = ; i <= n; i++) {
         for (int j = ; j <= n; j++) {
             for (int k = ; k <= n; k++) {
                 c.data[i][j] = (c.data[i][j] + 1ll * a.data[i][k] * b.data[k][j]) % m;
             }
         }
     }
 
     return c;
 }
 
 //矩阵加法
 matrix add(matrix a, matrix b) {
     for (int i = ; i <= n; i++) {
         for (int j = ; j <= n; j++) {
             a.data[i][j] = (a.data[i][j] + b.data[i][j])%m;
         }
     }
     return a;
 }
 
 //矩阵快速幂
 matrix quickpow(matrix a, int k) {
     matrix  c;
     memset(c.data, , sizeof(c.data));
     for (int i = ; i <= n; i++)
         c.data[i][i] = ;
     while (k) {
         if (k & ) c = mul(c, a);
         k >>= ;
         a = mul(a, a);
     }
     return c;
 }
 
 //正式计算 sum k
 matrix sum(matrix a, int k) {
     if (k == ) return a;
     matrix c;
     memset(c.data, , sizeof(c.data));
     for (int i = ; i <= n; i++)
         c.data[i][i] = ;
     c = add(c, quickpow(a, k >> ));
     c = mul(c, sum(a, k >> ));
     if (k & ) c = add(c, quickpow(a, k));
     return c;
 }
 
 /*
 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
 */
 
 int main()
 {
     matrix mat;
     cin >> n;
     cin >> k;
     cin >> m;
 
     for (int i = ; i <= n; i++) {
         for (int j = ; j <= n; j++) {
             cin >> mat.data[i][j];
         }
     }
 
     matrix ret = sum(mat, k);
 
     for (int i = ; i <= n; i++) {
         for (int j = ; j <= n; j++) {
             cout << ret.data[i][j] << " ";
         }
         cout << endl;
     }
 }