C - Searching Local Minimum

最新推荐文章于 2021-11-19 09:00:21 发布

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#二分法 #算法

这篇博客介绍了如何在给定一个未知的长度为n的数组时，通过互动查询找到局部最小值的索引。交互过程允许查询数组中任意位置的元素，不超过100次。博主提出利用二分法来解决问题，通过不断查询中间元素及其相邻元素，来确定局部最小值的位置。最终示例展示了二分法的实现过程。

题目

题目出处：https://codeforces.com/contest/1480/problem/C

C. Searching Local Minimum
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
This is an interactive problem.

Homer likes arrays a lot and he wants to play a game with you.

Homer has hidden from you a permutation a1,a2,…,an of integers 1 to n. You are asked to find any index k (1≤k≤n) which is a local minimum.

For an array a1,a2,…,an, an index i (1≤i≤n) is said to be a local minimum if ai<min{ai−1,ai+1}, where a0=an+1=+∞. An array is said to be a permutation of integers 1 to n, if it contains all integers from 1 to n exactly once.

Initially, you are only given the value of n without any other information about this permutation.

At each interactive step, you are allowed to choose any i (1≤i≤n) and make a query with it. As a response, you will be given the value of ai.

You are asked to find any index k which is a local minimum after at most 100 queries.

Interaction
You begin the interaction by reading an integer n (1≤n≤105) on a separate line.

To make a query on index i (1≤i≤n), you should output “? i” in a separate line. Then read the value of ai in a separate line. The number of the “?” queries is limited within 100.

When you find an index k (1≤k≤n) which is a local minimum, output “! k” in a separate line and terminate your program.

In case your query format is invalid, or you have made more than 100 “?” queries, you will receive Wrong Answer verdict.

After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

fflush(stdout) or cout.flush() in C++;
System.out.flush() in Java;
flush(output) in Pascal;
stdout.flush() in Python;
see documentation for other languages.
Hack Format

The first line of the hack should contain a single integer n (1≤n≤105).

The second line should contain n distinct integers a1,a2,…,an (1≤ai≤n).

Example
inputCopy
5
3
2
1
4
5
outputCopy
? 1
? 2
? 3
? 4
? 5
! 3

题意

给出一个n为数组大小，但你不能知道整体数组情况，你可以通过互动对数组进行查询，你可以选择任意i进行查询，并且回得到i位置的元素，问在不超过100次的查询中能够找出局部最小值的位置并且输出

思路

可以利用二分法进行解决，因为本题实际求的是 $a [m i d - 1] > a [m i d] 并且 a [m i d] < a [m i d + 1]$ 的 $a [m i d]$ 的 $m i d$ ,所以只要用二分法一直往小的方向查找，就会发现当 $r = = l$ 的时候就是局部最小值。故我们使用二分法的时候每一轮只要查询 $a [m i d], a [m i d + 1]$ 的时候即可，通过这两个数值的比较来确定查找方向。

代码实现

#include<iostream>
#include<cstdio>
using namespace std;
int a[100000];
int l,r;
void f(int x)
{
    if(x>=1&&x<=r)   //x没有超过查找范围
    {
      cout<<"? "<<x<<endl;  
      cin>>a[x];     
    }
}
int main()
{
    int n;
    cin>>n;
    
    l=1;
    r=n;
    a[0]=n+1;     //设置边界  边界值设置为数组里面最大值，因为在边界只需要比较一个值即可知道局部情况
    a[r+1]=n+1;   

    while(l<r)   //二分查找
    {
        int mid=(r+l)/2;   
        f(mid);         //查询  a[mid]的值
        f(mid+1);       //查询  a[mid+1]的值
        if(a[mid]>a[mid+1]) l=mid+1;   //a[mid]>a[mid+1]时，查找方向往右边走，即往小的方向走
        else r=mid;                    //反之，则想左边走
    }
    cout<<"! "<<r;      //输出结果
}