Training little cats
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
#include <cstdio> #include <cstring> #define ll long long struct matrix{ll m[105][105];}a; int n, k; ll m; // n:cat_num ,k:operation_num , m:operation_times matrix multiply(matrix x,matrix y) { matrix ans;//set an inside_ans to mark the solution memset(ans.m,0,sizeof(ans.m));//initialization the ans for(int i=1;i<=n;i++)for(int j=1;j<=n;j++) if(x.m[i][j])//if can operate for(int k=1;k<=n;k++)//just operate it ans.m[i][k]+=x.m[i][j]*y.m[j][k]; //matrix_multiply_above return ans;//at last ,return the ans } matrix quickmod(int p) { matrix ans;//set an inside_ans to mark the solution memset(ans.m,0,sizeof(ans.m));//initialization the ans for(int i=1;i<=n;i++)ans.m[i][i]=1;//initialization the map while(p)//if p !=0 , just go on { //bit operation if(p&1)//just as if(p%2==1) ans=multiply(ans,a);//renew the base_number p>>=1;//just as p/=2 a=multiply(a,a);//this operation lower the index amd renew the base_number } // if p==0, it means that we have operation p times already //this paragraph add the quilk_^_operation return ans;//at last ,return the ans } int main(){ll tmp; while(scanf("%d %lld %d",&n,&m,&k)!=EOF&&(n+m+k)) //if read case ,and n,m,k!=0,just go on { n++; char s[10];int x,y;memset(a.m,0,sizeof(a.m)); for(int i=1;i<=n;i++)a.m[i][i]=1;//initialization the map,one wide more //the more list is mark how many peanuts it have for(int i=0;i<k;i++){ scanf("%s",s);//read an order if(s[0]=='g')//case 1:take a peanut scanf("%d", &x),a.m[x][n]++; //read the i,and let it get a peanut if(s[0]=='e')//case 2:eat all the peanuts { scanf("%d",&x);//read the i for(int j=1;j<=n;j++)a.m[x][j]=0; //then set it's peanuts to 0 //this line turn 0, no operation it has } if(s[0]=='s')//case 3:change two cats' peanuts { scanf("%d %d",&x,&y);//read i and j for(int j=1;j<=n;j++)//then change their every operation tmp=a.m[x][j],a.m[x][j]=a.m[y][j],a.m[y][j]=tmp; } }matrix ans=quickmod(m);//set a out_side ans to inherit function_inside_ans for(int i=1;i<n;i++)printf("%lld ",ans.m[i][n]);printf(" "); //against every cat,printf it's peanuts_num }
program is above,the program is the best language