【CF39E】【博弈论】What Has Dirichlet Got to Do with That?
Description
You all know the Dirichlet principle, the point of which is that if n boxes have no less than n + 1 items, that leads to the existence of a box in which there are at least two items.
Having heard of that principle, but having not mastered the technique of logical thinking, 8 year olds Stas and Masha invented a game. There are a different boxes and b different items, and each turn a player can either add a new box or a new item. The player, after whose turn the number of ways of putting b items into a boxes becomes no less then a certain given number n, loses. All the boxes and items are considered to be different. Boxes may remain empty.
Who loses if both players play optimally and Stas's turn is first?
Input
The only input line has three integers a, b, n (1 ≤ a ≤ 10000, 1 ≤ b ≤ 30, 2 ≤ n ≤ 109) — the initial number of the boxes, the number of the items and the number which constrains the number of ways, respectively. Guaranteed that the initial number of ways is strictly less than n.
Output
Output "Stas" if Masha wins. Output "Masha" if Stas wins. In case of a draw, output "Missing".
Sample Input
2 2 10
Masha
5 5 16808
Masha
3 1 4
Stas
1 4 10
Missing
Hint
In the second example the initial number of ways is equal to 3125.
- If Stas increases the number of boxes, he will lose, as Masha may increase the number of boxes once more during her turn. After that any Stas's move will lead to defeat.
- But if Stas increases the number of items, then any Masha's move will be losing.
Source
1 /* 2 酒逢知己千杯少，话不投机半句多。 3 遥知湖上一樽酒，能忆天涯万里人。 4 */ 5 6 #include <set> 7 #include <map> 8 #include <cmath> 9 #include <cstdio> 10 #include <vector> 11 #include <cstring> 12 #include <cstdlib> 13 #include <iostream> 14 #include <algorithm> 15 #define LOCAL 16 const int MAXL = 20; 17 const long long MOD = 1000000007; 18 const int MAXK = 10000 + 10; 19 const int MAXN = 35000 + 10; 20 const int MAXM = 35; 21 using namespace std; 22 typedef long long LL; 23 int a , b , n , sg[MAXN][MAXM]; 24 25 LL pow(int a, int b){ 26 if (b == 0) return 1ll; 27 if (b == 1) return (LL)a; 28 LL tmp = pow(a, b / 2); 29 if (b % 2 == 0) return tmp * tmp; 30 else return (LL)tmp * tmp * (LL)a; 31 } 32 int dfs (int a , int b) { 33 if (sg[a][b] >= 0) return sg[a][b]; 34 if (pow(a, b) >= (LL)n) { 35 return sg[a][b] = 1; 36 } 37 int Ans = 0; 38 Ans |= !dfs(a + 1 , b);//只要有一个是1就能够获胜 39 Ans |= !dfs(a , b + 1); 40 return sg[a][b] = Ans; 41 } 42 int work_2(int a){ 43 int b = 1, turn = 0; 44 int up = (int)sqrt((double)n - 0.0000001);//处理上界 45 while (a <= up){ 46 if (sg[a][b + 1] == 0) return (turn == 0); 47 a++; 48 turn = turn ^ 1; 49 } 50 //剩下的就是判断奇偶性，看谁先到 51 turn = (turn + (n - 1 - a)) % 2; 52 return turn; 53 } 54 int work_1(int b){ 55 int a = 1; 56 if (pow(a, b) >= n) return 0; 57 else { 58 int turn = 0; 59 for (; !(pow(2, b) >= n); b++){ 60 if (sg[a + 1][b] == 0){ 61 if (turn == 0) return 1; 62 else return -1; 63 } 64 turn ^= 1; 65 } 66 return 0; 67 } 68 } 69 70 int main () { 71 72 memset(sg , -1 , sizeof (sg)); 73 scanf("%d%d%d", &a, &b, &n); 74 if (pow(a, b) >= n) { 75 puts ("Masha"); 76 return 0; 77 } 78 dfs(2, 2); 79 if (a != 1 && b != 1){ 80 if (sg[a][b] != 0) printf("Masha"); 81 else printf("Stas"); 82 return 0; 83 } 84 85 if (a == 1 && b == 1){ 86 int c = work_1(b + 1); 87 int d = work_2(a + 1); 88 if (c < 0 || d == 0) printf("Masha"); 89 else if (c == 0) printf("Missing"); 90 else printf("Stas"); 91 }else if (b == 1){ 92 if (work_2(a) != 0) printf("Masha"); 93 else printf("Stas"); 94 }else if (a == 1){ 95 int tmp = work_1(b); 96 if (tmp == 0) printf("Missing"); 97 else if (tmp == 1) printf("Masha"); 98 else printf("Stas"); 99 } 100 return 0; 101 }