样例一：
5 7
5 7 9 15 10

>> 9

样例二：
3 1
2 3 4

>> -1

解题思路

题目大概意思： 给你N个数， 还有一个数M， 让你求K， 进行M次减K操作， 令这N个数都小于等于0。

思路： 利用二分来快速找到K [N, sum(N)]， 找到后去验证。如果验证能行， 区间向左聚， 如果不能向右聚。

sum(N): N个数的和。

Python题解【二分查找】O(nlogn)

import math

class Solution:
    def check(self, target, nums, n):
        """
        验证是否可行
		"""
        count = 0
        for i in nums:
            if i <= target:
                count += 1
            else:
                count += math.ceil(i / target)
        return True if count <= n else False

    def run(self):
        m, n = map(int, input().split(" "))
        nums = list(map(int, input().split(" ")))
        l, r = 0, sum(nums)

        while l <= r:
            mid = (l + r) >> 1	# 除二
            if self.check(mid, nums, n):
                r = mid - 1	# 向左聚合
            else:
                l = mid + 1 # 向右聚合
        if l == -1 or l == sum(nums) + 1:
            print(-1)
            return
        print(l)


if __name__ == '__main__':
    Solution().run()

Java题解

import java.util.*;

public class Solution {
    private boolean check(int target, int[] nums, int n) {
        int count = 0;
        for (int i : nums) {
            if (i <= target) {
                count += 1;
            } else {
                count += Math.ceil((double)i / target);
            }
        }
        return count <= n;
    }

    public void run() {
        Scanner scanner = new Scanner(System.in);
        int m = scanner.nextInt();
        int n = scanner.nextInt();
        int[] nums = new int[m];
        for (int i = 0; i < m; i++) {
            nums[i] = scanner.nextInt();
        }
        int l = 0, r = Arrays.stream(nums).sum();

        while (l <= r) {
            int mid = (l + r) >> 1;
            if (check(mid, nums, n)) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        if (l == -1 || l == Arrays.stream(nums).sum() + 1) {
            System.out.println(-1);
            return;
        }
        System.out.println(l);
    }

    public static void main(String[] args) {
        new Solution().run();
    }
}

Javascript题解

function check(target, nums, n) {
    let count = 0;
    for (let i of nums) {
        if (i <= target) {
            count += 1;
        } else {
            count += Math.ceil(i / target);
        }
    }
    return count <= n;
}

function run() {
    const input = readline().split(' ').map(Number);
    const m = input[0];
    const n = input[1];
    const nums = readline().split(' ').map(Number);
    let l = 0, r = nums.reduce((a, b) => a + b);

    while (l <= r) {
        let mid = (l + r) >> 1;
        if (check(mid, nums, n)) {
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
    if (l === -1 || l === nums.reduce((a, b) => a + b) + 1) {
        console.log(-1);
        return;
    }
    console.log(l);
}

run();

Go题解

package main

import (
	"fmt"
	"math"
)

func check(target int, nums []int, n int) bool {
	count := 0
	for _, i := range nums {
		if i <= target {
			count += 1
		} else {
			count += int(math.Ceil(float64(i) / float64(target)))
		}
	}
	return count <= n
}

func main() {
	var m, n int
	fmt.Scanf("%d %d", &m, &n)
	nums := make([]int, m)
	for i := 0; i < m; i++ {
		fmt.Scan(&nums[i])
	}
	l, r := 0, 0
	for _, v := range nums {
		r += v
	}

	for l <= r {
		mid := (l + r) >> 1
		if check(mid, nums, n) {
			r = mid - 1
		} else {
			l = mid + 1
		}
	}
	if l == -1 || l == r+1 {
		fmt.Println(-1)
		return
	}
	fmt.Println(l)
}