Codeforces 39e what has Dirichlet got to do with that? Game + Memory search

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Question:

Given a box, B ball constant N (the ball and the box are different and there will be no identical items)

Set Way = put B balls in a box. If way> = N, the game ends.

Two people play games.

If the current turn is X

1. Select "X" to add a box or a ball.

2. If the number of methods> = N after adding the number, X fails.

If the first hand wins, the output is Masha. If the first hand wins, the output is Stas. If the first hand wins, the output is missing.

Ideas:

Memory-based search

If a ++ or B ++ is given at present, the current situation will fail if it is greater than or equal to n.

Transfer from the situation where no> = N is involved.

Note that if there is only one box and another box is added, it will be greater than or equal to N, so it is inevitable that only B will be added, so there is no solution.

If there is only one ball and another ball is no solution, then only a can be added, then the parity of N-A can be achieved to the current situation.

# Include <cstdio> # include <iostream> # include <string. h ># include <map> using namespace STD; # define ll long longll n, a, B; bool win (LL X, ll y) {// y balls in X boxes. The point is the Fail state ll TMP = 1; for (ll I = 1; I <= y; I ++) {TMP * = x; If (TMP> = N) return true;} return false;} Map <pair <LL, ll>, ll> MP; ll DFS (ll x, ll y) {// The status of this vertex if (MP. find (pair <LL, ll> (x, y ))! = MP. end () return MP [pair <LL, ll> (x, y)]; If (x = 1 & Win (2, y )) return MP [pair <LL, ll> (x, y)] =-1; if (y = 1 & Win (x, 2 )) {If (n-x) & 1) return MP [pair <LL, ll> (x, y)] = 0; return MP [pair <LL, ll> (x, y)] = 1;} If (WIN (x, y) return MP [pair <LL, ll> (x, y)] = 1; ll u = Win (x + 1, Y), V = Win (X, Y + 1); If (u = 1 & V = 1) return MP [pair <LL, ll> (x, y)] = 0; If (u = 0) u = DFS (x + 1, y ); if (V = 0) V = DFS (X, Y + 1); If (u = 0 | V = 0) Return MP [pair <LL, ll> (x, y)] = 1; if (u =-1 | V =-1) return MP [pair <LL, ll> (x, y)] =-1; return MP [pair <LL, ll> (x, y)] = 0;} int main () {ll, b; while (CIN> A> B> N) {MP. clear (); If (WIN (a + 1, B) & Win (A, B + 1) {puts ("Stas"); continue ;} if (B = 1 & Win (A, 2) {If (! (N-a) & 1) puts ("Masha"); else puts ("Stas"); continue;} ll TMP = DFS (A, B ); if (TMP <0) puts ("missing"); else TMP? Puts ("Masha"): puts ("Stas");} return 0 ;}