• CodeForces 753C Interactive Bulls and Cows (Hard)


    题意:。。。

    析:随机判断就即可,每次把不正确的删除,经过几次后就基本剩不下了。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #include <unordered_map>
    #include <unordered_set>
    #define debug() puts("++++");
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const LL LNF = 1LL << 60;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 100 + 5;
    const int mod = 2000;
    const int dr[] = {-1, 1, 0, 0};
    const int dc[] = {0, 0, 1, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c){
        return r >= 0 && r < n && c >= 0 && c < m;
    }
    bool use[15];
    
    P query(const string &s){
        cout << s << endl;
        int a, b;
        cin >> a >> b;
        return P(a, b);
    }
    
    void dfs(int d, string s, vector<string> &v){
        if(4 == d){
            v.push_back(s);
            return ;
        }
        for(int i = 0; i < 10; ++i)  if(!use[i]){
            use[i] = true;
            dfs(d+1, s + (char)('0' + i), v);
            use[i] = false;
        }
    }
    
    bool judge(const string &s1, const string &s2, P &p){
        int cnt1 = 0;
        for(int i = 0; i < 4; ++i)
            if(s1[i] == s2[i])  ++cnt1;
        int cnt2 = 0;
        for(int i = 0; i < 4; ++i)
            for(int j = 0; j < 4; ++j)
                if(s2[i] == s1[j]){ ++cnt2;  break;  }
        return cnt1 == p.first && (cnt2 - cnt1 == p.second);
    }
    
    vector<string> solve(P &p, vector<string> &v){
        vector<string> tmp;
        for(int i = 1; i < v.size(); ++i)
            if(judge(v[0], v[i], p))  tmp.push_back(v[i]);
        return tmp;
    }
    
    #include <ctime>
    int main(){
        fflush(stdout);
        srand(2333333);
        vector<string> v;
        memset(use, 0, sizeof use);
        dfs(0, "", v);
        while(true){
            random_shuffle(v.begin(), v.end());
            P tmp = query(v[0]);
            if(tmp.first == 4)  break;
            v = solve(tmp, v);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6286295.html
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