Radio Direction Finding

题目描述

This is an interactive problem.

Radio direction finding, also known as radio orienteering or radio fox hunting, is a sport that combines radio technology with outdoor navigation. Participants use specialized receivers to locate hidden radio transmitters, testing their sense of direction and radio operation skills.

A university offers a PE course in radio direction finding. During the final exam, the teacher takes the students to a local tourist attraction. The map of the attraction can be considered as a cycle of $n$ points ($n$ is an odd number), numbered clockwise as $1,2,\dots ,n$. The teacher has previously buried two transmitters at two different points and given you a radio receiver.

This receiver is special; each time it receives a signal, you need to manually press a button on the receiver, and then the receiver will tell you the sum of the shortest distances from these two transmitters to your current location (distance is defined as the number of edges traversed). The passing criterion for the final exam is to find the positions of these two transmitters using the receiver no more than $40$ times.

Now, please write an interactive program to successfully pass the final exam.

交互

First, read a positive integer $T$ ($1\le T\le 10^3$) from the standard input, indicating the number of test cases.

For each test case, first read a positive integer $n$ ($3\le n\le 10^9$, and $n$ is odd) from the standard input, indicating the number of points in the cycle.

Then, start the interaction process. Each time you interact, you can output the following two types of data to the standard output:

• ? x: Indicates that you are using the receiver at point $x$, and then your program needs to read an integer $dist$ from the standard input, indicating the sum of the shortest distances from the two transmitters to point $x$. You need to ensure that $1\le x\le n$.
  • For each set of data, the ? x operation should not exceed $40$ times. If you inquire more than the specified number of times, the interactor will immediately end and you will receive a “Wrong Answer” verdict.
• ! x y: Indicates your answer, i.e., the points where the two transmitters are located (order does not matter) are at $x$ and $y$. In this case:
  • If your output format is correct and the answer is correct, the interactor will immediately move on to the next test case;
  • Otherwise, if your output format is incorrect or the answer is incorrect, the interactor will immediately end and you will receive a “Wrong Answer” verdict.

提示

Each output needs to include a line break and flush the buffer, otherwise you may get unexpected results other than the correct answer. To flush the buffer, you can:

• For C or C++, use fflush(stdout) or cout.flush().
• For Java, use System.out.flush().
• For Python, use stdout.flush().

题目链接

Codeforces——传送门

思路&代码

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
typedef long long ll;int n; // 注意题目要求n为奇数，事实上n为偶数是无解的int normal(int a) // 将值转化为1~n之间的数，使代码更简洁
{a--;if (a < 0)a += n;a %= n;return a + 1;
}int query(int a) // 询问当前点到两个隐藏点的最短路长度之和
{cout << '?' << ' ' << normal(a) << '\n';cout.flush(); // 见题目所给提示int res;cin >> res;return res;
}int other(int a) // 已知一个隐藏点a的位置，返回另一个隐藏点b的位置
{int dis = query(a);              // 先找到两个隐藏点之间的距离if (query(normal(a - 1)) == dis) // 然后判断另一个点在逆时针方向还是顺时针方向return a - dis;elsereturn a + dis;
}void output(int a) // 输出答案
{int b = other(normal(a));cout << '!' << ' ' << normal(a) << ' ' << normal(b) << '\n';cout.flush(); // 见题目所给提示return;
}int main()
{ios::sync_with_stdio(0);cin.tie(0);int t;cin >> t;while (t--){cin >> n;int q1 = query(1), q2 = query(2); // 先询问1和2两个相邻点int dif = q2 - q1;switch (dif){case 0:{int l = 1, r = n / 2; // 整个环的一半中一定有询问的值与q1不相同的点while (l < r)         // 二分第一个询问值与q1不相同的点，这个点一定是隐藏点或者隐藏点的对称点{int mid = (l + r + 1) >> 1;if (query(mid + 1) == q1)l = mid;elser = mid - 1;}// l+1为二分得到的点if (q1 <= n / 2) // 若点1在两个隐藏点为端点的劣弧上，则二分得到的点为隐藏点output(l + 1);else // 否则二分得到的点为隐藏点的对称点{int x = l + 1 + (n + 1) / 2; // 再作一下对称即可得到隐藏点output(x);}}break;// 下面四种情况只需先求出两隐藏点距离dis，然后计算出一个隐藏点位置即可case 1:{int dis = n - q2;int x = 1 - (q1 - dis) / 2;output(x);}break;case -1:{int dis = n - q1;int x = 2 + (q2 - dis) / 2;output(x);}break;case 2:{int dis = query(2 - q2 / 2);int x = 2 - (q2 - dis) / 2;output(x);}break;case -2:{int dis = query(1 + q1 / 2);int x = 1 + (q1 - dis) / 2;output(x);}break;default:cout << "error\n"; // 正常输入输出下不会执行该行代码break;}}return 0;
}