CodeForces 778B - Bitwise Formula

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。

 #include <bits/stdc++.h>
 using namespace std;
 int n, m;
 map<string, int> mp;
 struct query{
     string f;
     int num;
     int a, op, b;
 }q[];
 void init()
 {
     cin >> n >> m;
     string s;
     mp["?"] = ;
     for (int i = ;  i <= n; i++)
     {
         cin >> s;
         mp[s] = i;
         cin >> s;
         cin >> s;
         if (s[] >= '' && s[] <= '')
         {
              q[i].op = ;
              q[i].f = s;
         }
         else
         {
             q[i].a = mp[s];
             cin >> s;
             if (s[] == 'A') q[i].op = ;
             if (s[] == 'O') q[i].op = ;
             if (s[] == 'X') q[i].op = ;
             cin >> s;
             q[i].b = mp[s];
         }
     }
 }
 int find(int x, int p)
 {
     int ret = ;
     q[].num = p;
     for (int i = ; i <= n; i++)
     {
         if (q[i].op == ) q[i].num = q[i].f[x]-'';
         else
         {
             int a = q[q[i].a].num;
             int b = q[q[i].b].num;
             if (q[i].op == ) q[i].num = a&b;
             if (q[i].op == ) q[i].num = a|b;
             if (q[i].op == ) q[i].num = a^b;
         }
         ret += q[i].num;
     }
     return ret;
 }
 void solve()
 {
     string ans1, ans2;
     for (int i = ; i < m; i++)
     {
         int x0 = find(i, );
         int x1 = find(i, );
         x0 <= x1 ? ans1 += '' : ans1 += '';
         x0 >= x1 ? ans2 += '' : ans2 += '';
     }
     cout << ans1 << endl << ans2 << endl;
 }
 int main()
 {
     init();
     solve();
 }