A题解 | #小美的排列询问#扫描一遍数组,判断a[i-1]和a[i]是否为x,y或y,x
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <unordered_set>
#include <vector>
#include <set>
#include <queue>
#include <iostream>
using namespace std;
void my_ans(){
long long i,n,m=0,c0,c1,t,x=0,y,mod = 100000000,pos;
cin>>n;
int a[n];
string ans ="No";
for(i=0;i<n;++i) cin>>a[i];
cin>>x>>y;
for(i=1;i<n;++i){
if (a[i] ==x && a[i-1] ==y){
ans = "Yes";break;
}
if (a[i] ==y && a[i-1] ==x){
ans = "Yes";break;
}
}
cout<<ans<<endl;
return;
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
long long t=1,i,j;
//scanf("%d",&t);
//cin>>t;
while(t>0){
--t;my_ans();
}
return 0;
}
// 64 位输出请用 printf("%lld")