题解 | #小美的排列询问#

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")