Codeforces Round #500 (Div. 2) BC

CodeForces	1013B	And
CodeForces	1013C	Photo of The Sky

B

可以发现只有一次与操作是有意义的，所以答案只有-1，0，1，2四种情况

 #include <bits/stdc++.h>
 #define show(a) cout << #a << " = " << a << endl;
 const int MOD = 1e9+;
 const int MAXN = ;
 const int INF = ;
 typedef long long ll;

 using namespace std;

 int main()
 {
     std::ios::sync_with_stdio(false);
     int n, x, ans = -;
     cin >> n >> x;
     int a[MAXN], point[MAXN];
     for (int i = ; i <= n; i++)
     {
         cin >> a[i];
     }
     sort(a+, a++n);
     for (int i = ; i <= n; i++)
     {
         if (a[i] == a[i-])
         {
             ans = ;
             break;
         }
     }
     if (ans == -)
     {
         int flg = ;
         for (int i = ; i <= n; i++)
             point[i] = a[i] & x;
         for (int i = ; i <= n; i++)
         {
             int s = lower_bound(a+, a++n, point[i]) - a;
             if (point[i] == a[s] && s != i)
             {
                 ans = , flg = ;
                 break;
             }
         }
         if (!flg)
         {
             sort(point+, point++n);
             for (int i = ; i <= n; i++)
             {
                 if (point[i] == point[i-])
                 {
                     ans = ;
                     break;
                 }
             }
         }
     }
     cout << ans << endl;
     return ;
 }

C

可以看成是红蓝染色，要求（红最大-红最小）*（蓝最大*蓝最小），排序后考虑两端颜色相同或不同，时间复杂度O（n）

 #include <bits/stdc++.h>
 #define show(a) cout << #a << " = " << a << endl;
 typedef long long ll;
 const int MOD = 1e9+;
 const int MAXN = ;
 const ll INF = 1e18;

 using namespace std;

 int main()
 {
     std::ios::sync_with_stdio(false);
     int n;
     cin >> n;
     ll num[n*+];
     for (int i = ; i <= *n; i++)
         cin >> num[i];
     sort(num+, num++*n);
     ll ans = INF;
     for (int i = ; i <= n; i++)
         ans = min(ans, (num[i+n-] - num[i]) * (num[*n] - num[]));
     ans = min(ans, (num[n*] - num[n+]) * (num[n] - num[]));
     cout << ans << endl;
     return ;
 }