poj 3735 Training little cats（矩阵连乘应用）_算法实验7:矩阵连乘 time limit: 1 sec memory limit: 128 mb -优快云博客

本文链接：https://blog.youkuaiyun.com/fp_hzq/article/details/8036835

本文探讨了一道涉及矩阵连乘原理的编程问题，旨在通过优化算法解决猫类动物的训练任务。具体而言，文章详细阐述了如何利用矩阵连乘的性质来简化操作序列，从而高效地分配和管理花生资源。此外，文章还提供了一个实例输入输出，展示了算法的实际应用。此题不仅考验了读者的编程技巧，也加深了对矩阵运算的理解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Training little cats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7423		Accepted: 1801

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

Source

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17), Facer

题目： http://poj.org/problem?id=3735

题意：有n只猫，和k个操作，每次可以给一只猫一个花生，或者让一只猫吃掉手上的所有花生，或者交换两只猫的花生，重复着k个操作m次

分析：本来是很简单的模拟题，不过m太大了，很容易联想到矩阵连乘。。。

然后就是如何转换的问题了，拿一颗花生，其实就是平移。。。。吃掉花生就把对应的行赋值0，交换的话，也就是交换两行

PS：我是不会矩阵的，看了别人的题解才知道平移，记下来了，看来高中落下的一大个坑啊

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int mm=111;
__int64 a[mm][mm],b[mm][mm],c[mm][mm];
int i,j,e,n,m,k;
char op[9];
void Multi(__int64 a[mm][mm],__int64 b[mm][mm])
{
    memset(c,0,sizeof(c));
    int i,j,k;
    for(i=1;i<=n+1;++i)
        for(k=1;k<=n+1;++k)
            if(a[i][k])for(j=1;j<=n+1;++j)
                if(b[k][j])c[i][j]+=a[i][k]*b[k][j];
    for(i=1;i<=n+1;++i)
        for(j=1;j<=n+1;++j)
            a[i][j]=c[i][j];
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&k),n+m+k)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=1;i<=n+1;++i)a[i][i]=b[i][i]=1;
        while(k--)
        {
            scanf("%s%d",op,&i);
            if(op[0]=='g')++a[i][n+1];
            if(op[0]=='e')
                for(j=0;j<=n+1;++j)a[i][j]=0;
            if(op[0]=='s')
            {
                scanf("%d",&j);
                for(e=1;e<=n+1;++e)
                    swap(a[i][e],a[j][e]);
            }
        }
        while(m)
        {
            if(m&1)Multi(b,a);
            Multi(a,a);
            m>>=1;
        }
        for(i=1;i<=n;++i)
            printf("%I64d%c",b[i][n+1],i<n?' ':'\n');
    }
    return 0;
}