Codeforces Round #435 (Div. 2) D. Mahmoud and Ehab and the binary string[二分]

Posted 2020-10-07 Meternal

 

题目：http://codeforces.com/problemset/problem/862/D

题意：交互题，询问15次以内Hamming distance，输出一个二进制串里任意一个0或1的位置

题解：极简单的二分，从最后一位先判断一个，然后二分 根据上次和本次的距离差是否等于二分长度判断在左端还是右端有需要寻找的值寻找另一个。

 

 1 #define _CRT_SECURE_NO_DEPRECATE
 2 #pragma comment(linker, "/STACK:102400000,102400000")
 3 #include<iostream>  
 4 #include<cstdio>  
 5 #include<fstream>  
 6 #include<iomanip>
 7 #include<algorithm>  
 8 #include<cmath>  
 9 #include<deque>  
10 #include<vector>
11 #include<bitset>
12 #include<queue>  
13 #include<string>  
14 #include<cstring>  
15 #include<map>  
16 #include<stack>  
17 #include<set>
18 #include<functional>
19 #define pii pair<int, int>
20 #define mod 1000000007
21 #define mp make_pair
22 #define pi acos(-1)
23 #define eps 0.00000001
24 #define mst(a,i) memset(a,i,sizeof(a))
25 #define all(n) n.begin(),n.end()
26 #define lson(x) ((x<<1))  
27 #define rson(x) ((x<<1)|1) 
28 #define inf 0x3f3f3f3f
29 typedef long long ll;
30 typedef unsigned long long ull;
31 using namespace std;
32 const int maxn = 1e3 + 5;
33 
34 string a;
35 int ans0, ans1;
36 int main()
37 {
38     ios::sync_with_stdio(false);
39     cin.tie(0); cout.tie(0);
40     int i, j, k, m, n;
41     cin >> n;
42     int mode;
43     a.append(n, \'0\');
44     int cas = 0, test;
45     cout << "? " << a << endl;
46     cout.flush();
47     cin >> cas;
48     a[n - 1] = \'1\';
49     cout << "? " << a << endl;
50     cout.flush();
51     cin >> test;
52     if (test == cas + 1) { ans0 = n; mode = 1; }
53     else { ans1 = n; mode = 0; }
54     if (mode == 0) { cas = n - cas; for (int i = 0; i < n; ++i)a[i] = \'1\'; }
55     else cas = test;
56     int l = 0, r = n - 2;
57     while (r - l > 1)
58     {
59         int mid = l + r >> 1;
60         for (int i = l; i <= mid; ++i)a[i] = mode + \'0\';
61         cout << "? " << a << endl;
62         cout.flush();
63         cin >> test;
64         int len = mid - l + 1;
65         if (test - len == cas) { l = mid + 1; cas = test; }
66         else { for (int i = l; i <= mid; ++i)a[i] = (mode ^ 1) + \'0\'; r = mid; }
67     }
68     a[l] = mode + \'0\';
69     cout << "? " << a << endl;
70     cout.flush();
71     cin >> test;
72     if (test == cas - 1)
73     {
74         if (mode == 1)ans1 = l+1;
75         else ans0 = l+1;
76     }
77     else
78     {
79         if (mode == 1)ans1 = r+1;
80         else ans0 = r+1;
81     }
82     cout << "! " << ans0 << " " << ans1 << endl;
83     cout.flush();
84     return 0;
85 }