Codeforces 39E-What Has Dirichlet Got to Do with That?_e. what has dirichlet got to do with that?-优快云博客

本文介绍了一个基于Dirichlet原理的游戏策略分析方法，通过暴力搜索算法来判断两名玩家在特定条件下是否能够获胜。游戏涉及不同数量的箱子和物品，目标是使物品放入箱子的组合方式超过给定阈值。

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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".

Examples
inputCopy
2 2 10
outputCopy
Masha
inputCopy
5 5 16808
outputCopy
Masha
inputCopy
3 1 4
outputCopy
Stas
inputCopy
1 4 10
outputCopy
Missing

Note

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.

题意：一开始给定a个箱子，b个球和常数n （球和箱子都是各不相同的，不会出现有一样的物品），若把b个球放到a个箱子中的方案数大于n则游戏结束。有2个人玩游戏，若当前轮到一个人时，这个人可以选择增加一个箱子或增加一个球，若增加完后方案数大于n则这个人输。若先手必胜，则输出 Masha ，若先手必败则输出 Stas ，若为平局则输出 Missing。

解题思路：暴力搜索。若当前给a++或b++都使方案数>=n则当前局势必败。若只有一个箱子，且再增加一个箱子就会使方案数大于n，那么只能增加a，那么只能增加b。

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <map>  
#include <set>
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>  

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a, b, n;

bool mypow(int k, int p)
{
	LL ans = 1, a = k;
	while (p)
	{
		if (p & 1) ans *= a;
		if (ans >= 1LL * n) return 0;
		a *= a;
		p /= 2;
	}
	return 1;
}

int dfs(int a, int b)
{
	if (a == 1)
	{
		if (!mypow(2, b)) return -1;
		if (!dfs(2, b)) return 1;
		return 1 - dfs(a, b + 1);
	}
	if (mypow(a, b + 1) && !dfs(a, b + 1)) return 1;
	if (mypow(a + 1, b) && !dfs(a + 1, b)) return 1;
	return 0;
}

int main()
{
	while (~scanf("%d %d %d", &a, &b, &n))
	{
		int ans = dfs(a, b);
		if (ans > 0) puts("Masha\n");
		else if (!ans) puts("Stas\n");
		else puts("Missing\n");
	}
	return 0;
}