CodeForces 1328B

For the given integer n (n>2) let’s write down all the strings of length n which contain n−2 letters ‘a’ and two letters ‘b’ in lexicographical (alphabetical) order.

Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1≤i≤n), that si<ti, and for any j (1≤j<i) sj=tj. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.

For example, if n=5 the strings are (the order does matter):

aaabb
aabab
aabba
abaab
ababa
abbaa
baaab
baaba
babaa
bbaaa
It is easy to show that such a list of strings will contain exactly n⋅(n−1)2 strings.

You are given n (n>2) and k (1≤k≤n⋅(n−1)2). Print the k-th string from the list.

Input

The input contains one or more test cases.

The first line contains one integer t (1≤t≤1e4) — the number of test cases in the test. Then t test cases follow.

Each test case is written on the the separate line containing two integers n and k (3≤n≤1e5,1≤k≤min(2⋅1e9,n⋅(n−1)2).

The sum of values n over all test cases in the test doesn’t exceed 105.

Output

For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).

Example Input

7
5 1
5 2
5 8
5 10
3 1
3 2
20 100

Output

aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa

题目大意：
输入一个t表示有t组测试样例，对于每组测试样例，给出n和k，表示有n长度的ab串，由n-2个a和2个b组成，k表示输出第k个字典序（可以看一下题面的样例）。

解题思路：
这道题n的范围是1e5，我刚开始用的next_permutation()函数，TLE了一发，发现这个是一道思维题，我们可以根据n和k的值推算出 b 的位置，然后把对应的位置替换成b即可。我们可以根据给出的样例得到规律，第一个b在倒数第二个位置的有1个，倒数第三个b位置的有3个，倒数第四个b有4个，我们可以得到第一个b的位置数列是 1 3 6 10… 这样，推一个公式：n(n-1)/2，n=2时倒数第二个b截止到第1个，n=3时倒数第三个b截止到第3个，就能得到 n的数即是倒数第n个b的截止数。这样枚举到1e5就能确定第一个b的位置了。再看k：k也有公式，n-(n(n-1)/2-k)-1，不过要先根据前面的公式推出n所处的区间，比如n如果等于2，则说明第一个b应该在倒数第二个位置，之后把n=2带进去，就等得到第二个b的位置。ps：求出来的值是表示倒数第几个位置。AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int _max=1e5+50;
using ll = long long;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		ll n,k;
		cin>>n>>k;
		ll p,q;
		for(ll i=1;i<_max;i++)
		  if(k<=(i*(i-1)/2))
		  {
			  p=i;//第一个b
			  break;
		  }
		q=p-(p*(p-1)/2-k)-1;//第二个b
		//cout<<p<<" "<<q<<endl;
		string s(n,'a');
		s[n-p]=s[n-q]='b';
		cout<<s<<endl;
	}
	//system("pause");
	return 0;
}