B. K-th Beautiful String

time limit per test ：1 seconds memory limit per test ：256 megabytes

input ：standard input output ：standard output

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):

1.aaabb
2.aabab
3.aabba
4.abaab
5.ababa
6.abbaa
7.baaab
8.baaba
9.babaa
10.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≤ $10^4$ ) — 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≤ $10^5$ ,1≤k≤min(2⋅ $10^9$ ,n⋅(n−1) / 2).

The sum of values n over all test cases in the test doesn't exceed $10^5$ .

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

output

aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa

题目大意

求由n - 2个字母‘a'与2个字母’b'组成的长度为n的字符串中字典序第k小的字符串

思路

确定两个b的位置即可

观察题目所给样例，第一个b在倒二位置的字符串有1个，倒三2个，倒四3个，倒五4个，所以第k个字符串我们需要找到一个pos，使得 $\frac{pos\cdot(pos + 1)}{2} \leqslant k$ ，那么第一个b的位置就是n - pos，第二个b的位置就是 $n - \frac{pos\cdot(pos - 1)}{2} + 1$

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e5 + 10;
 
int main()
{
    int t;
    cin >> t;
    while (t --){
        ll n, k;
        cin >> n >> k;
        ll x = sqrt(k);
        while (x * (x + 1) / 2 < k) x ++;//其实没必要这样找位置，直接O（n）暴力找，每次减n - i - 1就行,使用这种方式要开ll，不然会死循环
        ll pos1 = n - x;
        ll y = k - (x - 1) * x / 2;
        ll pos2 = n - y + 1;
        for (int i= 1; i <= n; ++i){
            if (i == pos1 || i == pos2) cout << 'b';
            else cout << 'a';
        }
        cout << endl;
 
    }
    return 0;
}

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【Codeforces Round #629 (Div. 3) / 1328B 】- K-th Beautiful String - 暴力