找m*n乘法表中第K小的。固定范围找第K大/小用二分值范围。二分的范围为给定乘法表取值范围[1, m * n]。每次对候选mid找乘法表中小于mid的个数，如果比K小说明mid还可以往前走。乘法表是:
1 2 3 ... n
2*1 2 * 2 2 * 3 ... 2 * n
...
m * 1 m * 2 .... m * n
因此可以遍历1...m看mid / i 能到哪。
/*
* @lc app=leetcode id=668 lang=cpp
*
* [668] Kth Smallest Number in Multiplication Table
*
* https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/description/
*
* algorithms
* Hard (42.78%)
* Likes: 387
* Dislikes: 19
* Total Accepted: 14.8K
* Total Submissions: 34.4K
* Testcase Example: '3\n3\n5'
*
*
* Nearly every one have used the Multiplication Table. But could you find out
* the k-th smallest number quickly from the multiplication table?
*
*
*
* Given the height m and the length n of a m * n Multiplication Table, and a
* positive integer k, you need to return the k-th smallest number in this
* table.
*
*
* Example 1:
*
* Input: m = 3, n = 3, k = 5
* Output:
* Explanation:
* The Multiplication Table:
* 1 2 3
* 2 4 6
* 3 6 9
*
* The 5-th smallest number is 3 (1, 2, 2, 3, 3).
*
*
*
*
* Example 2:
*
* Input: m = 2, n = 3, k = 6
* Output:
* Explanation:
* The Multiplication Table:
* 1 2 3
* 2 4 6
*
* The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6).
*
*
*
*
* Note:
*
* The m and n will be in the range [1, 30000].
* The k will be in the range [1, m * n]
*
*
*/
// @lc code=start
class Solution {
public:
int findKthNumber(int m, int n, int k) {
const auto LEX = [&](int x) {
int count = 0;
for (int i = 1; i <= m; ++i) {
count += min(n, x / i);
}
return count;
};
int start = 1, end = m * n;
while (start < end) {
int mid = start + (end - mid) / 2;
if (LEX(mid) < k) start = mid + 1;
else end = mid;
}
return start;
}
};
// @lc code=end