poj3735(矩阵快速幂)

最新推荐文章于 2021-09-11 19:55:36 发布

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最新推荐文章于 2021-09-11 19:55:36 发布

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分类专栏：数学数论

本文链接：https://blog.youkuaiyun.com/qq_25576697/article/details/77774458

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本文介绍了一种使用矩阵快速幂算法来模拟多只猫咪进行特定训练序列的问题。通过构建矩阵并利用快速幂运算，可以高效地计算出经过多次重复训练后每只猫咪所拥有的花生数量。

Training little cats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14625		Accepted: 3619

Description

Facer's pet cat just gave birth to a brood of little cats. Having considered the health of those lovely cats, Facer decides to make the cats to do some exercises. Facer has well designed a set of moves for his cats. He is now asking you to supervise the cats to do his exercises. Facer's great exercise for cats contains three different moves:
g i : Let the ith cat take a peanut.
e i : Let the ith cat eat all peanuts it have.
s i j : Let the ith cat and jth cat exchange their peanuts.
All the cats perform a sequence of these moves and must repeat it m times! Poor cats! Only Facer can come up with such embarrassing idea.
You have to determine the final number of peanuts each cat have, and directly give them the exact quantity in order to save them.

Input

The input file consists of multiple test cases, ending with three zeroes "0 0 0". For each test case, three integers n, m and k are given firstly, where n is the number of cats and k is the length of the move sequence. The following k lines describe the sequence.
(m≤1,000,000,000, n≤100, k≤100)

Output

For each test case, output n numbers in a single line, representing the numbers of peanuts the cats have.

Sample Input

Sample Output

2 0 1

题意：有n只猫一开始他们的值都是0，有三种操作，第i只猫值加1，第i只猫值变为0，dii只猫和第j只猫值交换，这样的操作选择几组，重复m次，求出最后每只猫的值。

思路：矩阵快速幂，就是建矩阵比较难，参考：http://blog.youkuaiyun.com/u013068502/article/details/38355621

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define ll long long
int n,m,k;
struct mat
{
	ll a[105][105];     //a[i][n]存放第i只猫有的豆子数 
};
mat mat_mul(mat x,mat y)  //矩阵相乘 
{  
    mat res;  
    memset(res.a,0,sizeof(res.a));  
    for(int i=0;i<=n;i++)  
        for(int j=0;j<=n;j++)  
        {
        	if(x.a[i][j])       //优化，判断这个点是否为0 
        	{
        		for(int k=0;k<=n;k++)
        		{
        			res.a[i][k]+=(x.a[i][j]*y.a[j][k]);
				}
			}
		}
    return res;  
}  
void solve()
{
	mat op,op0;
	memset(op.a,0,sizeof(op.a));
	for(int i=0;i<=n;i++)op.a[i][i]=1;
	op0=op;
	char s[2];
	int x,y;
	//先把做一次后的结果矩阵找出来 
	for(int i=0;i<k;i++)
	{
		scanf("%s",s);
		mat op1=op;
		if(s[0]=='g')
		{
			scanf("%d",&x);
			op1.a[x-1][n]=1;
		}
		else if(s[0]=='e')
		{
			scanf("%d",&x);
			op1.a[x-1][x-1]=0;
		}
		else if(s[0]=='s')
		{
			scanf("%d%d",&x,&y);
			op1.a[x-1][x-1]=0;
			op1.a[x-1][y-1]=1;
			op1.a[y-1][y-1]=0;
			op1.a[y-1][x-1]=1;
		}
		op0=mat_mul(op1,op0);
	}
	//矩阵快速幂 
	while(m)
	{
		if(m&1)op=mat_mul(op0,op);
		op0=mat_mul(op0,op0);
		m>>=1;
	}
	for(int i=0;i<n-1;i++)printf("%lld ",op.a[i][n]);
	printf("%lld\n",op.a[n-1][n]);
}
int main()
{
	while(~scanf("%d%d%d",&n,&m,&k)&&(n||m||k))
	{
		solve();
	}
}