B. Lost Number【CF交互题 暴力】

B. Lost Number【CF交互题 暴力】

This is an interactive problem. Remember to flush your output while

communicating with the testing program. You may use fflush(stdout) in

C++, system.out.flush() in Java, stdout.flush() in Python or

flush(output) in Pascal to flush the output. If you use some other

programming language, consult its documentation. You may also refer to

the guide on interactive problems:

https://codeforces.com/blog/entry/45307.

The jury guessed some array a consisting of 6 integers. There are 6

special numbers — 4, 8, 15, 16, 23, 42 — and each of these numbers

occurs in a exactly once (so, a is some permutation of these numbers).

You don’t know anything about their order, but you are allowed to ask

up to 4 queries. In each query, you may choose two indices i and j

(1≤i,j≤6, i and j are not necessarily distinct), and you will get the

value of ai⋅aj in return.

Can you guess the array a?

The array a is fixed beforehand in each test, the interaction program

doesn’t try to adapt to your queries.

Interaction

Before submitting the answer, you may ask up to 4 queries. To ask a

query, print one line in the following format: ? i j, where i and j

should be two integers such that 1≤i,j≤6. The line should be ended

with a line break character. After submitting a query, flush the

output and read the answer to your query — one line containing one

integer ai⋅aj. If you submit an incorrect query (or ask more than 4

queries), the answer to it will be one string 0. After receiving such

an answer, your program should terminate immediately — otherwise you

may receive verdict “Runtime error”, “Time limit exceeded” or some

other verdict instead of “Wrong answer”.

To give the answer, your program should print one line ! a1 a2 a3 a4

a5 a6 with a line break in the end. After that, it should flush the

output and terminate gracefully.

Example
inputCopy
16
64
345
672
outputCopy
? 1 1
? 2 2
? 3 5
? 4 6
! 4 8 15 16 23 42

Hint

If you want to submit a hack for this problem, your test should

contain exactly six space-separated integers a1, a2, …, a6. Each of 6

special numbers should occur exactly once in the test. The test should

be ended with a line break character.

思路如下

这是第一次写CF交互题，虽然没有做出来，不过看完题解之后这题是真的简单。

• 题意：一个数组有 4 ，8 ，15，16，32，42 这6个数子组成，且每个元素在该数组中仅出现一次，让你通过4 提问，得出这个 数组各个位置的元素是什么并输出出来。（每次提问方式如下： 设 i ， j 为所求数组的元素下标（i 与 j 可以相同），通过提问 i ，j 则系统会返回 i ，j 位置元素的乘积）
• 思路：固定提问位置 + 暴力枚举。

题解一（别人的）

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define ms(x, n) memset(x,n,sizeof(x));
typedef  long long LL;
const int inf = 1<<30;
const LL maxn = 1e5+10;

int main()
{
    int n[6] = {4,8,15,16,23,42}, ans[6];
    bool used[10];
    int a, b, c, d;
    ms(used, 0);

    printf("? 1 2\n");
    fflush(stdout);
    scanf("%d",&a);

    printf("? 2 3\n");
    fflush(stdout);
    scanf("%d",&b);

    printf("? 3 4\n");
    fflush(stdout);
    scanf("%d",&c);

    printf("? 4 5\n");
    fflush(stdout);
    scanf("%d",&d);

    for(int i = 0; i < 6; ++i){
        if(!used[i]){
            used[i] = true;
            for(int j = 0; j < 6; ++j){
                if(!used[j]){
                    used[j] = true;
                    for(int k = 0; k < 6; ++k){
                        if(!used[k]){
                            used[k] = true;
                            for(int l = 0; l < 6; ++l){
                                if(!used[l]){
                                    used[l] = true;
                                    for(int p = 0; p < 6; ++p){
                                        if(!used[p] && n[i]*n[j]==a&&n[j]*n[k]==b&&n[k]*n[l]==c&&n[l]*n[p]==d){
                                            used[p] = true;
                                            ans[0]=n[i], ans[1]=n[j], ans[2]=n[k], ans[3]=n[l], ans[4]=n[p];
                                            int t = 0;
                                            for(; t < 6; ++t)
                                                if(!used[t]) break;
                                            ans[5] = n[t];
                                            i=j=k=l=p=9; break;
                                        }
                                    }
                                    used[l] = false;
                                }
                            }
                            used[k] = false;
                        }

                    }
                    used[j] = false;
                }
            }
            used[i] = false;
        }
    }

    printf("! %d %d %d %d %d %d\n", ans[0], ans[1], ans[2], ans[3], ans[4], ans[5]);
	return 0;
}

题解二（改编后的题解）

经题解一改编后

#include<iostream>
using namespace std;
int res[7];
int mark[7];
int map[7] = {4,8,15,16,23,42};
int ar[7];

//暴力搜索回溯
int flag = 0;
void dfs(int k,int last_val)
{
    if(flag == 1)
        return;
    if(k == 5)
    {
        for(int i = 0; i < 6; i ++)
        {
            if(! mark[i])
            {
                res[k] = map[i];
                flag = 1;
                return;
            }
        }
    }

    for(int i = 0; i < 6; i ++)
    {
        if(k == 0 && !mark[i])
        {
            if(flag == 1)
                return;
            mark[i] = 1;
            res[k] = map[i];
            dfs(k + 1 , map[i]);
            mark[i] = 0;
        }
        else if(k > 0 && !mark[i] && ar[k] == last_val * map[i])
        {
            if(flag == 1)
                return;
            mark[i] = 1;
            res[k] = map[i];
            dfs(k + 1 , map[i]);
            mark[i] = 0;
        }
    }
}

int main()
{
    //freopen("test.txt","r",stdin);
    printf("? 1 2\n");
    fflush(stdout);
    scanf("%d",&ar[1]);

    printf("? 2 3\n");
    fflush(stdout);
    scanf("%d",&ar[2]);

    printf("? 3 4\n");
    fflush(stdout);
    scanf("%d",&ar[3]);

    printf("? 4 5\n");
    fflush(stdout);
    scanf("%d",&ar[4]);
    dfs(0 , 1);
    printf("! %d %d %d %d %d %d\n", res[0], res[1], res[2], res[3], res[4], res[5]);

    return 0;
}