codeforces 1167B Lost Numbers

传送门:https://codeforces.com/contest/1167/problem/B

题意:

交互题:现在你有6个数4, 8, 15, 16, 23, 42组成的某种组合,你可以询问系统四次,每次询问的格式为?i j

系统会回复你ai*aj的值,问你这个组合是哪种,有解且解是唯一的

题解:

交互题技巧,cin,cout可以自动刷新,而scanf,printf则需要手动用fflush(stdout)来刷新

根据观察我们可以发现,这六个数两两乘积不相等,所以a1*a2 a2*a3这样的解是唯一的,并且只有六个数,我们用next_permatation全排列函数可以得到结果

代码:

/**
 *        ┏┓    ┏┓
 *        ┏┛┗━━━━━━━┛┗━━━┓
 *        ┃       ┃  
 *        ┃   ━    ┃
 *        ┃ >   < ┃
 *        ┃       ┃
 *        ┃... ⌒ ...  ┃
 *        ┃       ┃
 *        ┗━┓   ┏━┛
 *          ┃   ┃ Code is far away from bug with the animal protecting          
 *          ┃   ┃   神兽保佑,代码无bug
 *          ┃   ┃           
 *          ┃   ┃        
 *          ┃   ┃
 *          ┃   ┃           
 *          ┃   ┗━━━┓
 *          ┃       ┣┓
 *          ┃       ┏┛
 *          ┗┓┓┏━┳┓┏┛
 *           ┃┫┫ ┃┫┫
 *           ┗┻┛ ┗┻┛
 */
// warm heart, wagging tail,and a smile just for you!
//
//                            _ooOoo_
//                           o8888888o
//                           88" . "88
//                           (| -_- |)
//                           O  =  /O
//                        ____/`---'\____
//                      .'  |     |//  `.
//                     /  |||  :  |||//  
//                    /  _||||| -:- |||||-  
//                    |   | \  -  /// |   |
//                    | \_|  ''---/''  |   |
//                      .-\__  `-`  ___/-. /
//                  ___`. .'  /--.--  `. . __
//               ."" '<  `.___\_<|>_/___.'  >'"".
//              | | :  `- \`.;` _ /`;.`/ - ` : | |
//                 `-.   \_ __ /__ _/   .-` /  /
//         ======`-.____`-.___\_____/___.-`____.-'======
//                            `=---='
//        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
//                     佛祖保佑      永无BUG
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]
"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
"

const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;

int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    vector<int> v {4, 8, 15, 16, 23, 42};
    int a, b, c, d;
    cout << "? 1 2" << endl;
    cin >> a;
    cout << "? 2 3" << endl;
    cin >> b;
    cout << "? 4 5" << endl;
    cin >> c;
    cout << "? 5 6" << endl;
    cin >> d;
    do {
        if(v[0]*v[1] == a && v[1]*v[2] == b && v[3]*v[4] == c && v[4]*v[5] == d) {
            cout<<"! ";
            for(auto it : v) {
                cout<<it<<" ";
            }
            break;
        }
    } while (next_permutation(v.begin(), v.end()));
    cout<<endl;
    return 0;
}
每一个不曾刷题的日子 都是对生命的辜负 从弱小到强大,需要一段时间的沉淀,就是现在了 ~buerdepepeqi
原文地址:https://www.cnblogs.com/buerdepepeqi/p/10876263.html