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UVA 11149 - Power of Matrix

题目链接

题意：给定一个n*n的矩阵A和k，求∑kiAi

思路：利用倍增去搞，∑kiAi=(1+Ak/2)∑k/2iAi，不断二分即可

代码：

#include <cstdio>
#include <cstring>

const int N = 45;

int n, k;

struct mat {
	int v[N][N];
	mat() {memset(v, 0, sizeof(v));}
 	mat operator * (mat c) {
	 	mat ans;
	 	for (int i = 0; i < n; i++) {
	 		for (int j = 0; j < n; j++) {
	 			for (int k = 0; k < n; k++) {
	 				ans.v[i][j] = (ans.v[i][j] + v[i][k] * c.v[k][j]) % 10;
     			}
    		}
		}
		return ans;
  	}
  	mat operator + (mat c) {
  		mat ans;
  		for (int i = 0; i < n; i++)
  			for (int j = 0; j < n; j++)
  				ans.v[i][j] = (v[i][j] + c.v[i][j]) % 10;
		return ans;
   	}
} A;

mat pow_mod(mat x, int k) {
	mat ans;
	for (int i = 0; i < n; i++) ans.v[i][i] = 1;
	while (k) {
		if (k&1) ans = ans * x;
		x = x * x;
		k >>= 1;
 	}
 	return ans;
}

mat solve(mat x, int k) {
	if (k == 1) return x;
	mat ans;
	for (int i = 0; i < n; i++) ans.v[i][i] = 1;
	if (k == 0) return ans;
	ans = (ans + pow_mod(x, k>>1))* solve(x, k>>1);
	if (k&1) ans = ans + pow_mod(x, k);
	return ans;
}

int main() {
	while (~scanf("%d%d", &n, &k) && n) {
 		for (int i = 0; i < n; i++)
   			for (int j = 0; j < n; j++) {
      			scanf("%d", &A.v[i][j]);
      			A.v[i][j] %= 10;
   			}
  	    A = solve(A, k);
  	    for (int i = 0; i < n; i++)
  	    	for (int j = 0; j < n; j++)
  	    		printf("%d%c", A.v[i][j], (j == n - 1 ? '\n' : ' '));
		printf("\n");
 	}
	return 0;
}

UVA 11149 - Power of Matrix(矩阵倍增),布布扣,bubuko.com

UVA 11149 - Power of Matrix(矩阵倍增)

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原文地址：http://blog.csdn.net/accelerator_/article/details/37994841