给定正整数
n
,
k
n,k
n,k，考虑所有由
n
?
2
n-2
n?2 个 a 和
2
2
2 个 b 构成的字符串，输出其中字典序第
k
k
k 小的。
多组数据， T ≤ 1 0 4 T\le10^4 T≤104， 3 ≤ n ≤ 1 0 5 3\le n\le 10^5 3≤n≤105， k ≤ min ? ( 2 × 1 0 9 , n × ( n ? 1 ) 2 ) k\le\min(2\times10^9,\frac{n\times(n-1)}2) k≤min(2×109,2n×(n?1)?)， ∑ n ≤ 1 0 5 \sum n\le10^5 ∑n≤105。
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 \le i \le n $ ), that $ s_i < t_i $ , and for any $ j $ ( $ 1 \le j < i $ ) $ s_j = t_j $ . 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):
It is easy to show that such a list of strings will contain exactly $ \frac{n \cdot (n-1)}{2} $ strings.
You are given $ n $ ( $ n > 2 $ ) and $ k $ ( $ 1 \le k \le \frac{n \cdot (n-1)}{2} $ ). Print the $ k $ -th string from the list.
The input contains one or more test cases.
The first line contains one integer $ t $ ( $ 1 \le t \le 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 \le n \le 10^5, 1 \le k \le \min(2\cdot10^9, \frac{n \cdot (n-1)}{2}) $ .
The sum of values $ n $ over all test cases in the test doesn’t exceed $ 10^5 $ .
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).
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
int findfirst(int n,int *second)
{
int i = 1;
int sum = 0;
for (;;)
{
for (int m=1;m<=i;m++ )
{
sum++;
if (sum == n)
{
*second = m;
return i + 1;
}
}
i++;
}
}
//时间超限的程序。下面没有用
#include<stdio.h>
int main(void)
{
int t;
scanf("%d", &t);
while (t--)
{
int n = 0, k = 0;
char arr[1000000] = { 0 };
scanf("%d %d", &n, &k);
int first = 1;//第1个b位置
int s = 0;
int j = 1;
while (s < k) {
s += j;
j++;
}
first = j;
int second = first - 1 - (s - k);
for (int i = 1; i <= n - first; i++)
{
printf("a");
}
printf("b");
for (int i = 1; i <= first - second-1; i++)
{
printf("a");
}
printf("b");
for (int i = 1; i <= second - 1; i++)
{
printf("a");
}
printf("\n");
}
return 0;
}
不会二分硬是做成数学题。。。。