UVA 11149 Power of Matrix 快速幂

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122094#problem/G

Power of Matrix

Time Limit:3000MS
Memory Limit:0KB
#### 问题描述
> 给你一个矩阵A，求A+A^2+A^3+...+A^k
#### 输入
> Input consists of no more than 20 test cases. The first line for each case contains two positive integers n
> (≤ 40) and k (≤ 1000000). This is followed by n lines, each containing n non-negative integers, giving
> the matrix A.
> Input is terminated by a case where n = 0. This case need NOT be processed.

输出

For each case, your program should compute the matrix A + A2 + A3 + . . . + Ak

. Since the values may

be very large, you only need to print their last digit. Print a blank line after each case.

样例

sample input

3 2

0 2 0

0 0 2

0 0 0

0 0

sample output

0 2 4

0 0 2

0 0 0

题解

A+A^2+...Ak=(I+A^{(n/2))(A+...+A}(n/2))=...

一直递推下去，深度只有logn，只要logn次快速幂求A^x。

代码

WA四次的代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn = 55;
const int mod = 10;
typedef int LL;

struct Matrix {
	LL mat[maxn][maxn];
	Matrix() { memset(mat, 0, sizeof(mat)); }
	friend Matrix operator *(const Matrix& A, const Matrix& B);
	friend Matrix operator +(const Matrix &A,const Matrix &B);
	friend Matrix operator ^(Matrix A, int n);
};

Matrix I;

Matrix operator +(const Matrix& A, const Matrix& B) {
	Matrix ret;
	for (int i = 0; i < maxn; i++) {
		for (int j = 0; j < maxn; j++) {
			ret.mat[i][j] = (A.mat[i][j] + B.mat[i][j])%mod;
		}
	}
	return ret;
}

Matrix operator *(const Matrix& A, const Matrix& B) {
	Matrix ret;
	for (int i = 0; i < maxn; i++) {
		for (int j = 0; j < maxn; j++) {
			for (int k = 0; k < maxn; k++) {
				ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k] * B.mat[k][j]) % mod;
			}
		}
	}
	return ret;
}

Matrix operator ^(Matrix A, int n) {
	Matrix ret=I;
	while (n) {
		if (n & 1) ret = ret*A;
		A = A*A;
		n /= 2;
	}
	return ret;
}

Matrix solve(Matrix A, int n) {
	if (!n) return I;
	if (n == 1) return A;
	//这里要加括号！！！  ret=I+(A^(n/2))!!!
	Matrix ret = I + A ^ (n / 2);
	ret = ret*solve(A, n / 2);
	//这里也要加！！！！ ret=ret+(A^n)!!!
	if (n % 2) ret = ret + A^n;
	return ret;
}

int n, k;

int main() {
	for (int i = 0; i < maxn; i++) I.mat[i][i] = 1;
	while (scanf("%d%d", &n, &k) == 2 && n) {
		Matrix A;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				scanf("%d", &A.mat[i][j]);
				A.mat[i][j] %= mod;
			}
		}
		Matrix ans=solve(A, k);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n-1; j++) {
				printf("%d ",ans.mat[i][j]);
			}
			printf("%d\n", ans.mat[i][n - 1]);
		}
		printf("\n");
	}
	return 0;
}

保险一些的写法：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn = 55;
const int mod = 10;
typedef int LL;

struct Matrix {
	LL mat[maxn][maxn];
	Matrix() { memset(mat, 0, sizeof(mat)); }
	friend Matrix operator *(const Matrix& A, const Matrix& B);
	friend Matrix operator +(const Matrix &A,const Matrix &B);
	friend Matrix pow(Matrix A, int n);
};

Matrix I;

Matrix operator +(const Matrix& A, const Matrix& B) {
	Matrix ret;
	for (int i = 0; i < maxn; i++) {
		for (int j = 0; j < maxn; j++) {
			ret.mat[i][j] = (A.mat[i][j] + B.mat[i][j])%mod;
		}
	}
	return ret;
}

Matrix operator *(const Matrix& A, const Matrix& B) {
	Matrix ret;
	for (int i = 0; i < maxn; i++) {
		for (int j = 0; j < maxn; j++) {
			for (int k = 0; k < maxn; k++) {
				ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k] * B.mat[k][j]) % mod;
			}
		}
	}
	return ret;
}

Matrix pow(Matrix A, int n) {
	Matrix ret=I;
	while (n) {
		if (n & 1) ret = ret*A;
		A = A*A;
		n /= 2;
	}
	return ret;
}

Matrix solve(Matrix A, int n) {
	if (!n) return I;
	if (n == 1) return A;
	Matrix ret = I + pow(A,n/2);
	ret = ret*solve(A, n / 2);
	if (n % 2) ret = ret + pow(A,n);
	return ret;
}

int n, k;

int main() {
	for (int i = 0; i < maxn; i++) I.mat[i][i] = 1;
	while (scanf("%d%d", &n, &k) == 2 && n) {
		Matrix A;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				scanf("%d", &A.mat[i][j]);
				A.mat[i][j] %= mod;
			}
		}
		Matrix ans=solve(A, k);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n-1; j++) {
				printf("%d ",ans.mat[i][j]);
			}
			printf("%d\n", ans.mat[i][n - 1]);
		}
		printf("\n");
	}
	return 0;
}