POJ-3735 Training little cats(矩阵快速幂)

Training little cats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13586		Accepted: 3362

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

矩阵快速幂

程序：

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <queue>
 6 #include <stack>
 7 #include <vector>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 typedef long long LL;
12 const int MAX=105;
13 LL n,m,K;
14 struct Mat{
15     LL x,y;
16     LL mat[MAX][MAX];
17     Mat(){
18         x=0,y=0;
19         memset(mat,0,sizeof(mat));
20     }
21     Mat operator * (const Mat &cc) {
22         Mat zt;
23         zt.x=x;
24         zt.y=cc.y;
25         int i,j,k;
26         for (i=1;i<=x;i++)
27          for (k=1;k<=cc.x;k++)
28           if (mat[i][k])
29            for (j=1;j<=cc.y;j++)
30             zt.mat[i][j]+=mat[i][k]*cc.mat[k][j];
31         return zt;
32     }
33 }a,b;
34 Mat ksm(Mat zt,LL k){
35     LL i,j;
36     Mat an;
37     an.x=zt.x,an.y=zt.y;
38     for (i=1;i<=an.x;i++)
39      for (j=1;j<=an.y;j++)
40       an.mat[i][j]=(i==j);
41     while (k)
42     {if (k%2==1)
43       an=an*zt;
44      k/=2;
45      zt=zt*zt;
46     }
47     return an;
48 }
49 int main(){
50     freopen ("cats.in","r",stdin);
51     freopen ("cats.out","w",stdout);
52     LL i,j;
53     LL zt1,zt2;
54     while (1)
55     {scanf("%lld%lld%lld",&n,&m,&K);
56      if (n==0 && m==0 && K==0)
57       break;
58      b.x=n+1,b.y=n+1;
59      for (i=1;i<=b.x;i++)
60       for (j=1;j<=b.y;j++)
61        b.mat[i][j]=(i==j);
62      char s[MAX];
63      a.x=n+1,a.y=1;
64      memset(a.mat,0,sizeof(a.mat));
65      a.mat[n+1][1]=1;
66      for (i=1;i<=K;i++)
67      {scanf("%s",s+1);
68       if (s[1]=='g')
69       {scanf("%lld\n",&zt1);
70        b.mat[zt1][n+1]++;
71       }
72       if (s[1]=='e')
73       {scanf("%lld\n",&zt1);
74        for (j=1;j<=n+1;j++)
75         b.mat[zt1][j]=0;
76       }
77       if (s[1]=='s')
78       {scanf("%lld%lld\n",&zt1,&zt2);
79        for (j=1;j<=n+1;j++)
80         swap(b.mat[zt1][j],b.mat[zt2][j]);
81       }
82      }
83      b=ksm(b,m);
84      a=b*a;
85      for (i=1;i<=n;i++)
86       printf("%lld ",a.mat[i][1]);
87      printf("\n");
88     }
89     return 0;
90 }

posted on 2016-10-13 23:36 珍珠鸟阅读(313) 评论(0) 编辑收藏举报