【每日算法Day 83】邻居小孩一年级就会的乘法表，你会吗？

题目链接

LeetCode 668. 乘法表中第k小的数[1]

示例1

输入：
m = 3, n = 3, k = 5
输出：
3
解释：
乘法表:
1 2 3
2 4 6
3 6 9
第5小的数字是 3 (1, 2, 2, 3, 3).

示例2

输入：
m = 2, n = 3, k = 6
输出：
6
解释：
乘法表:
1 2 3
2 4 6
第6小的数字是 6 (1, 2, 2, 3, 4, 6).

题解

代码

二分法（c++）

class Solution {
public:
    int findKthNumber(int m, int n, int k) {
        int l = 1, r = m*n;
        while (l < r) {
            int mid = l+((r-l)>>1);
            if (enough(mid, m, n, k)) r = mid;
            else l = mid+1;
        }
        return l;
    }

    bool enough(int x, int m, int n, int k) {
        int cnt = 0;
        for (int i = 1; i <= m; ++i) {
            cnt += x/i<n?x/i:n;
        }
        return cnt >= k;
    }
};

二分法+优化（c++）

class Solution {
public:
    int findKthNumber(int m, int n, int k) {
        int l = 1, r = k;
        while (l < r) {
            int mid = l+((r-l)>>1);
            if (enough(mid, m<mid?m:mid, n<mid?n:mid, k)) r = mid;
            else l = mid+1;
        }
        return l;
    }

    bool enough(int x, int m, int n, int k) {
        int cnt = n*(x/n), d = 0;
        for (int i = (x/n)+1; i <= m; i = d+1) {
            d = x/(x/i);
            cnt += (x/i)*((d<m?d:m)-i+1);
        }
        return cnt >= k;
    }
};

二分法（python）

class Solution:
    def findKthNumber(self, m: int, n: int, k: int) -> int:
        def enough(x, m, n, k):
            cnt = 0
            for i in range(1, m+1):
                cnt += x//i if x//i<n else n
            return cnt >= k

        l, r = 1, m*n
        while l < r:
            mid = l+((r-l)>>1)
            if enough(mid, m, n, k): r = mid
            else: l = mid+1
        return l

二分法+优化（python）

class Solution:
    def findKthNumber(self, m: int, n: int, k: int) -> int:
        def enough(x, m, n, k):
            cnt, i, d = n*(x//n), x//n+1, 0
            while i <= m:
                d = x//(x//i)
                cnt += (x//i)*((d if d<m else m)-i+1)
                i = d+1
            return cnt >= k

        l, r = 1, k
        while l < r:
            mid = l+((r-l)>>1)
            if enough(mid, m if m<mid else mid, n if n<mid else mid, k): r = mid
            else: l = mid+1
        return l

参考资料

[1]

LeetCode 668. 乘法表中第k小的数: ​​https://leetcode-cn.com/problems/kth-smallest-number-in-multiplication-table/​​