题目描述
Slime has a sequence of positive integers a1,a2,…,an.
In one operation Orac can choose an arbitrary subsegment [l…r] of this sequence and replace all values al,al+1,…,ar to the value of median of {al,al+1,…,ar}.
In this problem, for the integer multiset s, the median of s is equal to the ⌊|s|+1/2⌋-th smallest number in it. For example, the median of {1,4,4,6,5} is 4, and the median of {1,7,5,8} is 5.
Slime wants Orac to make a1=a2=…=an=k using these operations.
Orac thinks that it is impossible, and he does not want to waste his time, so he decided to ask you if it is possible to satisfy the Slime’s requirement, he may ask you these questions several times.
Input
The first line of the input is a single integer t: the number of queries.
The first line of each query contains two integers n (1≤n≤100000) and k (1≤k≤109), the second line contains n positive integers a1,a2,…,an (1≤ai≤109)
The total sum of n is at most 100000.
Output
The output should contain t lines. The i-th line should be equal to ‘yes’ if it is possible to make all integers k in some number of operations or ‘no’, otherwise. You can print each letter in lowercase or uppercase.
Example
input
5
5 3
1 5 2 6 1
1 6
6
3 2
1 2 3
4 3
3 1 2 3
10 3
1 2 3 4 5 6 7 8 9 10
output
no
yes
yes
no
yes
Note
In the first query, Orac can’t turn all elements into 3.
In the second query, a1=6 is already satisfied.
In the third query, Orac can select the complete array and turn all elements into 2.
In the fourth query, Orac can’t turn all elements into 3.
In the fifth query, Orac can select [1,6] at first and then select [2,10].
题目大意
一个数组里一段数可以都变成他的中位数，问整个序列中的数能不能都变成k。可以变化很多次。
题目分析
能成功变化需要满足的条件有：
- a[]中有k这个数。
- 有a[i]>=k，a[j]>=k，且|j-i|<=2或n==1。
若不满足这两个条件，则输出no。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <unordered_set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=1e5+5;
int a[N];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,k;
scanf("%d%d",&n,&k);
bool flag=true;
for(int i=1;i<=n;i++) //判断a[]中是否有k
{
scanf("%d",&a[i]);
if(a[i]==k) flag=false;
}
if(flag) {puts("no"); continue;}
for(int i=1;i<n;i++) //判断条件二
if(a[i]>=k)
{
if(a[i+1]>=k||(a[i+2]>=k&&i+2<=n))
{
flag=true;
break;
}
}
if(flag||n==1) puts("yes");
else puts("no");
}
return 0;
}