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
分析：
因为所有数字都是m位的，高位的权重大于低位 ，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。
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n, m; 4 map<string, int> mp; 5 struct query{ 6 string f; 7 int num; 8 int a, op, b; 9 }q[5005]; 10 void init() 11 { 12 cin >> n >> m; 13 string s; 14 mp["?"] = 0; 15 for (int i = 1; i <= n; i++) 16 { 17 cin >> s; 18 mp[s] = i; 19 cin >> s; 20 cin >> s; 21 if (s[0] >= '0' && s[0] <= '9') 22 { 23 q[i].op = 0; 24 q[i].f = s; 25 } 26 else 27 { 28 q[i].a = mp[s]; 29 cin >> s; 30 if (s[0] == 'A') q[i].op = 1; 31 if (s[0] == 'O') q[i].op = 2; 32 if (s[0] == 'X') q[i].op = 3; 33 cin >> s; 34 q[i].b = mp[s]; 35 } 36 } 37 } 38 int find(int x, int p) 39 { 40 int ret = 0; 41 q[0].num = p; 42 for (int i = 1; i <= n; i++) 43 { 44 if (q[i].op == 0) q[i].num = q[i].f[x]-'0'; 45 else 46 { 47 int a = q[q[i].a].num; 48 int b = q[q[i].b].num; 49 if (q[i].op == 1) q[i].num = a&b; 50 if (q[i].op == 2) q[i].num = a|b; 51 if (q[i].op == 3) q[i].num = a^b; 52 } 53 ret += q[i].num; 54 } 55 return ret; 56 } 57 void solve() 58 { 59 string ans1, ans2; 60 for (int i = 0; i < m; i++) 61 { 62 int x0 = find(i, 0); 63 int x1 = find(i, 1); 64 x0 <= x1 ? ans1 += '0' : ans1 += '1'; 65 x0 >= x1 ? ans2 += '0' : ans2 += '1'; 66 } 67 cout << ans1 << endl << ans2 << endl; 68 } 69 int main() 70 { 71 init(); 72 solve(); 73 }