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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
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 5 using namespace std; 6 int n, t, m; 7 struct mat { 8 long long a[102][102]; 9 mat() { 10 memset(a, 0, sizeof(a)); 11 } 12 mat operator *(const mat &o) const { 13 mat t; 14 for(int i = 0; i <= n; i++) { 15 for(int j = 0; j <= n; j++) 16 for(int k = 0; k <= n; k++) { 17 if(a[i][k]){ 18 t.a[i][j] = (t.a[i][j] + a[i][k]*o.a[k][j]); 19 } 20 } 21 } 22 return t; 23 } 24 }; 25 26 int main() { 27 28 while(~scanf("%d%d%d" , &n, &m, &t)) { 29 if(n == 0 && m == 0 && t == 0) break; 30 mat a, b, c; 31 for(int i = 0; i <= n; i++) a.a[i][i] = 1; 32 for(int i = 0; i <= n; i++) b.a[i][i] = 1; 33 for(int i = 0; i <= n; i++) c.a[i][i] = 1; 34 // c.a[0][0] = 1; 35 while(t--){ 36 char s[20]; 37 int x, y; 38 scanf("%s", s); 39 if(s[0] == ‘s‘){ 40 scanf("%d%d", &x, &y); 41 for(int i = 0; i <= n; i++){ 42 swap(b.a[i][x], b.a[i][y]); 43 } 44 }else if(s[0] == ‘g‘){ 45 scanf("%d", &x); 46 b.a[0][x]++; 47 }else { 48 scanf("%d", &x); 49 for(int i = 0; i <= n; i++) b.a[i][x] = 0; 50 } 51 } 52 53 while(m > 0) { 54 if(m&1) { 55 a = a*b; 56 m--; 57 } 58 m >>= 1; 59 b = b*b; 60 } 61 a = c*a; 62 for(int i = 1; i <= n; i++){ 63 if(i == 1) printf("%lld", a.a[0][i]); 64 else printf(" %lld", a.a[0][i]); 65 } 66 printf("\n"); 67 } 68 return 0; 69 }
poj 3735 Training little cats矩阵快速幂
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原文地址：http://www.cnblogs.com/cshg/p/5936416.html
