POJ 3233：Matrix Power Series 矩阵快速幂 乘积

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 18450		Accepted: 7802

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

题意很简单，就是矩阵相乘，然后求和。自己做的时候快速幂，发现快速幂竟然还是TLE。

不知道怎么搞，看了网上的代码，发现这个求和的深搜sum2很经典，充分利用偶数求和，假设是求1到6的和，先将6除以2，求1到3的和，然后对1到3的和 乘以3就是4到6的和，再一相加就是1到6的和。这段代码的思想很巧妙，很喜欢。以后求1到n的和时候可以用得上~

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

struct matrix {
	int m[35][35];
};

int n, mod;
long long ko;
matrix no;

matrix mu(matrix no1, matrix no2)
{
	matrix t;
	memset(t.m, 0, sizeof(t.m));

	int i, j, k;
	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= n; j++)
		{
			for (k = 1; k <= n; k++)
			{
				t.m[i][j] += no1.m[i][k] * no2.m[k][j];
				if (t.m[i][j] >= mod)
				{
					t.m[i][j] %= mod;
				}
			}
		}
	}
	return t;

}

matrix multi(matrix no, long long x)
{
	matrix b;
	memset(b.m, 0, sizeof(b.m));
	int i;
	for (i = 1; i <= n; i++)
	{
		b.m[i][i] = 1;
	}
	while (x)
	{
		if (x & 1) b = mu(b, no);
		x = x >> 1;
		no = mu(no, no);
	}
	return b;
}

matrix add(matrix no1, matrix no2)
{
	matrix t;

	int i, j;

	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= n; j++)
		{
			t.m[i][j] = no1.m[i][j] + no2.m[i][j];
			if (t.m[i][j] >= mod)
			{
				t.m[i][j] %= mod;
			}
		}
	}
	return t;
}

matrix sum2(long long i)//假设i为7
{
	if (i == 1)return no;
	if (i & 1)
		return add(multi(no, i), sum2(i - 1));//7+6+...
	else
	{
		long long k = i >> 1;//3
		matrix s = sum2(k);//1 2 3
		return add(s, mu(s, multi(no, k)));1 2 3 + 4 5 6
	}

}

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	int i, j;
	cin >> n >> ko >> mod;

	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= n; j++)
		{
			scanf("%d", &no.m[i][j]);
			if (no.m[i][j] >= mod)
			{
				no.m[i][j] %= mod;
			}
		}
	}
	no = sum2(ko);
	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= n; j++)
		{
			if (j == 1)
				cout << no.m[i][j];
			else
				cout << " " << no.m[i][j];
		}
		cout << endl;
	}
	//system("pause");
	return 0;
}

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