POJ - 3233 Matrix Power Series （矩阵等比二分求和）

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

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

题意：求矩阵总和

思路：矩阵快速幂取模，和等比数列矩阵求和，这里说一下怎么二分求矩阵的等比序列和，设矩阵为A，次数为k

             设sum(k) = A^1 + A^2 + A^3 + ..... + A^K,那么为了能够二分递归下去我们首先拆出个( A^1 + A^2 + ... + A^(k/2) ,可以很容易得到：sum(k) = sum(k/2) * (A^(k/2) + 1)

  在代码里是如果这个次数是奇数的话会少算一个A^k,所以要记得加上

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
typedef long long ll;
using namespace std;
const int maxn = 32;

int m, n;
struct Matrix {
	int v[maxn][maxn];
	Matrix() {}
	Matrix(int x) {
		init();
		for (int i = 0; i < maxn; i++) 
			v[i][i] = x;
	}
	void init() {
		memset(v, 0, sizeof(v));
	}
	Matrix operator *(Matrix const &b) const {
		Matrix c;
		c.init();
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				for (int k = 0; k < n; k++)
					c.v[i][j] = (c.v[i][j] + (v[i][k]*b.v[k][j])) % m;
		return c;
	}
	Matrix operator ^(int b) {
		Matrix a = *this, res(1);
		while (b) {
			if (b & 1)
				res = res * a;
			a = a * a;
			b >>= 1;
		}
		return res;
	}
} u(1);

Matrix Add(Matrix a, Matrix b) {
	for (int i = 0; i < maxn; i++)
		for (int j = 0; j < maxn; j++)
			a.v[i][j] = (a.v[i][j]+b.v[i][j]) % m;
	return a;
}

Matrix BinarySum(Matrix a, int n) {
	if (n == 1)
		return a;
	if (n & 1)
		return Add(BinarySum(a, n-1), a^n);
	else return BinarySum(a, n>>1) * Add(u, a^(n>>1));
}

int main() {
	int k;
	Matrix a, ans;
	while (scanf("%d%d%d", &n, &k, &m) != EOF) {
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				scanf("%d", &a.v[i][j]);
		ans = BinarySum(a, k);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				printf("%d%c", ans.v[i][j], (j==n-1)?'\n':' ');
	}
	return 0;
}