poj 3233 Matrix Power Series(矩阵幂的和,矩阵快速幂)
2016-11-10 12:01
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此题要求的是A+A^2+A^3...+A^k,当然逐个求出作和n^3*k复杂度肯定是要超时的。 所以可以构造一个矩阵Sk = I(单位矩阵) + A + A^2 + A^3...+ A^k-1,
Sk = Sk-1 + Ak-1.
所以就可以构造出矩阵快速幂的形式:
![](https://img-blog.csdn.net/20161110115757054?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center)
每个方格就是一个n*n的矩阵。 要求A+A^2+A^3...+A^k就是要求Sk+1 - I,求出矩阵的k+1次幂后的2n*2n的大矩阵的右上角的小矩阵即使Sk+1,然后减去单位矩阵即是答案。
代码:
Matrix Power Series
Description
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
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
Sample Output
Source
POJ Monthly--2007.06.03, Huang, Jinsong
Sk = Sk-1 + Ak-1.
所以就可以构造出矩阵快速幂的形式:
每个方格就是一个n*n的矩阵。 要求A+A^2+A^3...+A^k就是要求Sk+1 - I,求出矩阵的k+1次幂后的2n*2n的大矩阵的右上角的小矩阵即使Sk+1,然后减去单位矩阵即是答案。
代码:
#include<iostream> #include<cstdio> #include<cstring> using namespace std; typedef long long ll; ll n, k, mod; struct node { ll s[100][100]; void init(void){ memset(s, 0, sizeof(s)); } }; void show(node a, int t) { for(int i = 0; i < t; i++) for(int j = 0; j < t; j++) { printf("%d ", a.s[i][j]); if(j == t-1) printf("\n"); } } node mul(node a, node b) { node t; t.init(); for(int i = 0; i < 2*n; i++) for(int j = 0; j < 2*n; j++) for(int k = 0; k < 2*n; k++) t.s[i][j] = (t.s[i][j]+a.s[i][k]*b.s[k][j])%mod; return t; } node mt_pow(node p, int m) { node q; q.init(); for(int i = 0; i < 2*n; i++) q.s[i][i] = 1; while(m) { if(m&1) q = mul(q, p); p = mul(p, p); m /= 2; } return q; } int main(void) { while(~scanf("%lld%lld%lld", &n, &k, &mod)) { node base; base.init(); for(int i = 0; i < n; i++) for(int j = 0; j < 2*n; j ++) if(i == j || i+n == j) base.s[i][j] = 1; for(int i = n; i < 2*n; i++) for(int j = n; j < 2*n; j++) scanf("%d", &base.s[i][j]); // show(base, 2*n); node ans = mt_pow(base, k+1); // show(ans, 2*n); for(int i = 0; i < n; i++) { for(int j = n; j < 2*n; j++) { if(j-n) printf(" "); if(i+n == j) printf("%lld", (ans.s[i][j]-1+mod)%mod); else printf("%lld", ans.s[i][j]%mod); } printf("\n"); } } return 0; }
Matrix Power Series
Time Limit: 3000MS | Memory Limit: 131072K | |
Total Submissions: 21455 | Accepted: 8997 |
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
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
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