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")