[sg function]split game (2020 Jiangxi university student programming competition)

At the beginning there was a $n\times m$ Paper of , Then as the game goes on , You will get multiple pieces of mesh paper of different sizes , In this way, you can remember a piece of paper as a game , Each game constitutes the sum of a game . Each of these games is a directed graph game , But we have to find a situation where we can't act by ourselves . obviously $1\ast 2$ And $1\ast 3$（$2\ast 1$ and $3\ast 1$ Empathy ） Cannot split again , And other rectangles can be divided into one of the two , Therefore, these two situations are selected as the inevitable situation .

For one $N\times M$ Rectangular grid paper , We can enumerate how to act , Then get two sub Games , For two sub Games $SG$ Value is XOR , Then you get the corresponding after cutting SG value . For all the situations after cutting $SG$ Value for $mex$ operation （ Record all values as a set , Returns the smallest number that does not exist in the set ） You can get the current paper SG value .

Be careful to avoid $1\times 1$ The situation of

#include <bits/stdc++.h>
#define int long long
using namespace std;

#define win cout << "Alice" << endl
#define lose cout << "Bob" << endl

const int N = 200;
int sg[N][N];

void getsg(){
    
    memset(sg, -1, sizeof(sg));
    sg[1][1] = sg[1][2] = sg[2][1] = sg[1][3] = sg[3][1] = 0;
    for(int i = 1; i <= N; i++){
    
        for(int j = 1; j <= N; j++){
    
            set<int> st;
            for(int k = 1; k < i; k++){
    
                int lx = k, ly = j;
                int rx = i - k, ry = j;
                if((lx == 1 && ly == 1) || (rx == 1 && ry == 1)) continue;
                st.insert(sg[lx][ly] ^ sg[rx][ry]);
            }
            for(int k = 1; k < j; k++){
    
                int lx = i, ly = k;
                int rx = i, ry = j - k;
                if((lx == 1 && ly == 1) || (rx == 1 && ry == 1)) continue;
                st.insert(sg[lx][ly] ^ sg[rx][ry]);
            }
            int mex = 0;
            while(st.count(mex)) mex++;
            sg[i][j] = mex;
        }
    }
}


inline void solve(){
    
    getsg(); 
    int n, m;
    while(cin >> n >> m){
    
        if(sg[n][m]) win;
        else lose;
    }

}

signed main(){
    
    solve();
    return 0;
}