题目链接：这里
题意：选择一个 m 位的二进制数字，总分为 n 个算式的答案之和。问得到最低分和最高分分别应该取哪个二进制数字
解法：因为所有数字都是m位的，因为高位的权重大于地位 ，我们就从高到低考虑 ans 的每一位是取 0 还是取 1，统计该位的权重（即n个式子该位结果之和）即可。

//CF 779E

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5005;
map <string, int> mp;
struct Q{
    string s;
    int num, a, b, op;
    Q(){}
    Q(string s, int num, int a, int b, int op) : s(s), num(num), a(a), b(b), op(op) {}
}q[maxn];
int n, m;
int cal(int x, int p){
    q[0].num = p;
    int ans = 0;
    for(int i = 1; i <= n; i++){
        if(q[i].op == 0) q[i].num = q[i].s[x] - '0';
        else{
            int a = q[q[i].a].num, b = q[q[i].b].num;
            if(q[i].op == 1) q[i].num = a&b;
            if(q[i].op == 2) q[i].num = a|b;
            if(q[i].op == 3) q[i].num = a^b;
        }
        ans += q[i].num;
    }
    return ans;
}
int main()
{
    cin >> n >> m;
    string s;
    mp["?"] = 0;
    for(int i = 1; i <= n; i++){
        cin >> s;
        mp[s] = i;
        cin >> s;
        cin >> s;
        if(isdigit(s[0])){
            q[i].op = 0;
            q[i].s = s;
        }
        else{
            q[i].a = mp[s];
            cin >> s;
            if(s[0] == 'A') q[i].op = 1;
            if(s[0] == 'O') q[i].op = 2;
            if(s[0] == 'X') q[i].op = 3;
            cin >> s;
            q[i].b = mp[s];
        }
    }
    string ans1 = "", ans2 = "";
    for(int i = 0; i < m; i++){
        int cnt1 = cal(i, 0);
        int cnt2 = cal(i, 1);
        if(cnt1 <= cnt2) ans1 += '0'; else ans1 += '1';
        if(cnt1 >= cnt2) ans2 += '0'; else ans2 += '1';
    }
    cout << ans1 << endl;
    cout<< ans2 << endl;
}