POJ-3735 Training little cats

Training little cats Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14166 Accepted: 3511

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

3 1 6 g 1 g 2 g 2 s 1 2 g 3 e 2 0 0 0

Sample Output

2 0 1

Source

#include <iostream> #include <stdio.h> #include <string.h> using namespace std; typedef long long ll; #define maxn 105 int n,m,k; struct Matrix{ ll mat[maxn][maxn]; void clear(){ memset(mat,0,sizeof(mat)); } void unit(){ clear(); for(int i=0;i<maxn;i++){ mat[i][i]=1; } } }M,T; Matrix operator *(const Matrix &a,const Matrix &b){ Matrix tmp; tmp.clear(); for(int k=0;k<=n;k++){ for(int i=0;i<=n;i++){ if(a.mat[i][k]) for(int j=0;j<=n;j++){ tmp.mat[i][j]+=a.mat[i][k]*b.mat[k][j]; } } } return tmp; } Matrix operator ^(Matrix x,int n){ Matrix tmp; tmp.unit(); while(n){ if(n&1) tmp=tmp*x; x=x*x; n>>=1; } return tmp; } void init(){ M.clear(); M.mat[0][0]=1; T.unit(); } int main(){ char s[5]; int a,b; while(~scanf("%d%d%d",&n,&m,&k)&&(n+m+k)){ init(); for(int i=0;i<k;i++){ scanf("%s",s); if(s[0]=='g'){ scanf("%d",&a); T.mat[0][a]++; } else if(s[0]=='e'){ scanf("%d",&a); for(int i=0;i<=n;i++){ T.mat[i][a]=0; } } else{ scanf("%d%d",&a,&b); for(int i=0;i<=n;i++){ swap(T.mat[i][a],T.mat[i][b]); } } } Matrix ans=M*(T^m); for(int i=1;i<=n;i++){ printf("%I64d ",ans.mat[0][i]); } printf("\n"); } return 0; }