我的思路是枚举每一个拼音看看有没有在文章里出现。
可是拼音的表没打完，只有 $20$ 分。

#include<iostream>
#include<algorithm>
#define int long long
#define f(i,j,n) for(register int i=j;i<=n;++i)
using namespace std;
const int mod=998244353;
int n,T,ge;
string pinyin[11451]={"ba","bo","bi","bu","pa","po","pi","pu","ma","mo","me","mi","mu","fa","fo","fu","da","de","di","du","ta","te","ti","tu","na","ne","ni","nu","la","lo","le","li","lu","ga","ge","gu","ka","ke","ku","ha","he","hu","ji","ju","qi","qu","xi","xu","zha","zhe","zhi","zhu","cha","che","chi","chu","sha","she","shi","shu","re","ri","ru","za","ze","zi","zu","ca","ce","ci","cu","sa","se","si","su","ya","yo","ye","yi","yu","wa","wo","wu"};
string s[1010];
inline void solve(){
	f(i,1,n){
		f(j,0,ge){
			if(s[i]==pinyin[j]){
//				cout<<s[i]<<"=="<<pinyin[j]<<"\n";
				cout<<"Pinyin\n";
				return;
			}
		}
	}
	cout<<"English\n";
	return;
}
inline void print(){
	f(i,1,n){
		f(j,0,s[i].size()-1){
			cout<<s[i][j]<<" ";
		}
		cout<<"\n";
	}
	return;
}
signed main(){
	cin>>T;
	for(ge=0;pinyin[ge]!="\0";++ge);
	--ge;//找到一共几个拼音 
//	cout<<(pinyin[1145]=="\0")<<"\n";//空的以"\0" 
	while(T--){
		cin>>n;
		f(i,1,n)cin>>s[i];
//		print();
		solve();
	}
	return 0;
}
/*
in
2
14 zhe ge ti mu qi shi bi ni xiang xiang de yao jian dan
6 this problem has a simple solution
out
Pinyin
English
*/