題目鏈接：這里 題意：選擇一個 m 位的二進制數字，總分為 n 個算式的答案之和。問得到最低分和最高分分別應該取哪個二進制數字 解法：因為所有數字都是m位的，因為高位的權重大于地位 ，我們就從高到低考慮 ans 的每一位是取 0 還是取 1，統計該位的權重（即n個式子該位結果之和）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;}新聞熱點
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