POJ 3745 Training Little cats (Matrix operation + matrix fast power) __algorithm

POJ 3745 Training Little cats (Matrix operation + matrix fast power)
Time limit:2000ms Memory limit:65536kb

Description
Facer ' s pet cat just gave birth to a brood little of cats. has considered the health of those lovely cats, Facer decides to make the cats to do some. Facer has designed a set of moves for his cats. He's now asking you to supervise the cats to does 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.
The Cats perform a sequence of these moves and must repeat it m times! Poor cats! Only facer can come up and such embarrassing idea.
You are have to determine the final number of peanuts each cat have and directly give them the exact e 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 was 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
PKU Campus 2009 (POJ monthly contest–2009.05.17), Facer

First notice that the number of repetitions is very large, so there must be a mathematical algorithm.
Look at the operation, all linear operation, so you can use the linear transformation of vectors to handle these operations, the only problem is that there is a plus one of the operation, so the peanut quantity vector augmentation, add a non second dimension can be.
Then for X=[1,X1,X2,..., Xn]t,xi as the first cat peanut number x=[1,x_1,x_2,..., x_n]^t,x_i as the first cat peanut number
X=am[1,0,0...,0]t x=a^m[1,0,0...,0]^t
A can be obtained by converting a simple linear transformation into a matrix operation.
Use a fast power again.

Accepted    3840kB  812ms   1824 B  g++

#include <stdio.h> const int n=101;

const int POWER=50;

typedef long Long data_t;
data_t A[n][n];
data_t Ap[power][n][n];
/* x ' =ax x[0] is always 1, x[i] is the Peanuts cat I owns.
*/void CopyTo (data_t a[n][n],data_t b[n][n],int N);
void Mul (data_t a[n][n],data_t b[n][n],int N);
void Power (data_t a[n][n],int n,int m);

void Init (data_t a[n][n],int N);
    int main () {//freopen ("In.txt", "R", stdin);
    char cmd;
    int n,m,k,x,y,temp;
        while (scanf ("%d%d%d\n", &n,&m,&k) &&n) {init (a,n);
            while (k--) {scanf ("%c", &cmd);
                if (cmd== ' E ') {scanf ("%d\n", &x); 
            for (int i=0;i<=n;i++) a[x][i]=0;
                else if (cmd== ' s ') {scanf ("%d%d\n", &x,&y);
                    for (int i=0;i<=n;i++) {temp=a[x][i]; A[x][i]=a[y][i]; 
                A[y][i]=temp;
                } else if (cmd== ' G ') {scanf ("%d\n", &x); 
            a[x][0]++;
        } power (A,N,M); for (int i=1;i<=n;i++) printf ("%lld%c", a[i][0],i==n?)
    \ n ': ');
return 0;
    } void Mul (data_t a[n][n],data_t b[n][n],int N) {static data_t a0[n][n];
    for (int i=0;i<=n;i++) for (int j=0;j<=n;j++) A0[I][J]=A[I][J];
            for (int i=0;i<=n;i++) for (int j=0;j<=n;j++) {a[i][j]=0;
        for (int k=0;k<=n;k++) A[I][J]+=A0[I][K]*B[K][J];
} return;
    } void Power (data_t a[n][n],int n,int m) {int = =;
    data_t power=1;
    CopyTo (A,ap[0],n);
        while (power< (data_t) m) {power*=2;
        p++;
        CopyTo (Ap[p-1],ap[p],n);
    Mul (Ap[p],ap[p-1],n);
    Init (a,n); for (int i=0;i<=p;i++) if (m& (1<<i)) Mul (a,ap[i],n);
Return
            } void CopyTo (data_t a[n][n],data_t b[n][n],int N) {for (int i=0;i<=n;i++) for (int j=0;j<=n;j++)
    B[I][J]=A[I][J];
Return } void Init (data_t a[n][n],int N) {for (int i=0;i<=n;i++) for (int j=i+1;j<=n;j++) A[I][J]
    =a[j][i]= (data_t) 0;
    for (int i=0;i<=n;i++) a[i][i]= (data_t) 1;
Return }