Codeforces 39E What have Dirichlet Got to does with? Games + Memory Search

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Serie A champions:

Specific a box B ball does not change n (Ball and box are not the same, kind of goods)

Set way = Number of methods to put the B ball into a box, the end of the game if it >= N

There are 2 of people playing games.

If the current turn is X

1. x Select Add a box or add a ball

2, if the number of methods after adding >=n X failed

If the initiator wins, then output Masha, if the initiator will be defeated output Stas, if the draw is output Missing

Ideas:

Memory Search

If the current to a++ or b++ are >=n, the current situation will be defeated

From the situation that will not >=n.

Note that if there is only one box. and add a box will be >=n, then must only add B, then there must be no solution

If there is only one ball, and then add a ball to the case of no solution, then only can add a, then according to the parity of n-a can be 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) {//X box y ball This point is a must-fail 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 state of this point 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 A, 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"):p UTS ("Stas"); } return 0;}

Codeforces 39E What have Dirichlet Got to does with? Games + Memory Search