codeforces 1167B 交互ez

没咋接触过交互题

题意:给了6个数分别是4,8,15,16,23,42。会由coder询问四次,询问的方式为“? a b”。

  每次询问会由oj返回1个数,这个数为a,b的积。然后让你求出这六个数的顺序。

询问为cout

得到oj返回的数为cin

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '

#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
int q[7] = {4,8,15,16,23,42};
int main()
{
    int a,b,c,d;
    cout<<"?"<<sp<<1<<sp<<2<<endl;
    cin>>a;
    cout<<"?"<<sp<<1<<sp<<3<<endl;
    cin>>b;    
    cout<<"?"<<sp<<4<<sp<<5<<endl;
    cin>>c;
    cout<<"?"<<sp<<4<<sp<<6<<endl;
    cin>>d;
    do{
      if((a==q[1]*q[0])&&(b==q[0]*q[2])&&(c==q[3]*q[4])&&(d==q[3]*q[5])){
          cout<<"! ";
          rep(i,0,5){
              cout<<q[i]<<sp;
          }
          //cout<<endl;
      }     
    }while(next_permutation(q,q+6));
}
原文地址:https://www.cnblogs.com/jrfr/p/13298004.html