aaaaaaaaaaA heH heH nuN

aaaaaaaaaaA heH heH nuN

The question ：

Specify that the prefix of a string fragrance is nunhehheh, The suffix is a number a; Given a n , To find a n String of aromatic subsequences .

Ideas ：

nunhehheh + a * x, Yes 2^x - 1 Valid subsequences

nunhehhe hhhha , Yes p individual h, Then there are p Valid subsequences

So in the last one a Add one in front h It can form 2 ^ x Valid substrings

for example haha == 4 haaha == 8

It corresponds directly to 2 Base number , So we directly enumerate how many binary bits there are 1, Then in the front a Back plus h And the corresponding binary bit plus h( namely a And a Space between ) You can construct

AC Code ：

#include <iostream>
#include <algorithm>
#include <bitset>
#include <string.h>

using namespace std;
int s[50];

void solve(){
	int n;cin >> n;
	bitset<33>q(n);
	memset(s,0,sizeof s);
	if(q.test(0)) s[1] ++;
	for(int i = 1;i <= 31;i ++)
		if(q.test(i)) s[i]++,s[1]++;
	cout << "nunhehhe";
	for(int i = 31;i > 0;i --){
		for(int j = 0;j < s[i];j ++)
		 	cout << "h";
		cout << "a"; 
	} 
	cout << endl;
	return; 
}

int main(){
	int t;cin >> t;
	while(t --){
		solve();
	}
	return 0;
}