华为OD机试-日志限流-2022Q4 A卷-Py/Java/JS

某软件系统会在运行过程中持续产生日志，系统每天运行N单位时间，运行期间每单位时间产生的日志条数保行在数组 records中。records[i]表示第i单位时间内产生日志条数。
由于系统磁盘空间限制，每天可记录保存的日志总数上限为total条。

        如果一天产生的日志总条数大于total，则需要对当天内每单位时间产生的日志条数进行限流后保存，请计算每单位时间最大可保存日志条数limit，以确保当天保存的总日志条数不超过total。
        1：对于单位时间内产生日志条数不超过limit的日志全部记录保存:
        2：对于单位时间内产生日志条数超过limit的日志，则只记录保存limit条日志;

        如果一天产生的日志条数总和小干等于total，则不需要启动限流机制，result为-1。请返回result的最大值或者-1。
输入描述
第一行为系统某一天运行的单位时间数N.1<=N<=10^5
第二行为表示这一天每单位时间产生的日志数量的数组records，0<= records[i]<= 10^5第三行为系统一天可以保存的总日志条数total。1 <= total <= 10^9

输出描述
每单位时间内最大可保存的日志条数limit，如果不需要启动限流机制，返回-1。

示例1：输入输出示例仅供调试，后台判题数据一般不包含示例
输入

6
3 3 8 7 10 15

40
输出

9

Java 代码

import java.util.Scanner;
import java.util.*;
import java.util.stream.Collectors;
 
public class Main { 
    public static void main(String[] args) {
        //处理输入
        Scanner in=new Scanner(System.in); 
        int N = in.nextInt();
        int[] records = new int[N];
        long single_total = 0;
        for (int i = 0; i < N; i++) {
            records[i] = in.nextInt();
            single_total += records[i];
        }
        int total = in.nextInt();
 
        // 一天产生的日志总条数小于等于total
        if(single_total <= total) {
            System.out.println(-1);
            return;
        } else {
            Arrays.sort(records);
 
            //二分法初始化
            int left = total / N;
            int right = records[N - 1];
            
            int result = left;
            while (right > left+1) {
                int mid = (right + left) / 2;
            
                int temp_total = 0;
                for (int i=0; i total) {
                    right = mid;
                } else if (temp_total < total) {
                    left = mid;
                    result = mid;
                } else {
                    System.out.println(mid);
                    return;
                }
            }
        
            System.out.println(result);
            return;
        }
        
    }
 
    
}

Python代码

import functools
import collections
import math
from itertools import combinations
from re import match
 
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
 
#并查集模板
class UF:
    def __init__(self, n=0):
        self.count = n
        self.item = [0 for x in range(n+1)]
        for i in range(n):
            self.item[i] = i
    def find(self, x):
        if (x != self.item[x]):
            self.item[x] = self.find(self.item[x])
            return 0
        return x
    
 
    def union_connect(self, x, y):
        x_item = self.find(x)
        y_item = self.find(y)
    
        if (x_item != y_item):
            self.item[y_item] = x_item
            self.count-=1
 
 
 
# 处理输入
N = int(input())
records = [int(x) for x in input().split(" ")]
total = int(input())
single_total = sum(records)
 
find_flag = False
 
# 一天产生的日志总条数小于等于total
if(single_total <= total):
    print(-1)
else:
    records.sort()
    #二分法初始化
    left = total / N
    right = records[N - 1]
    result = left
    while (right > left+1):
        mid = int((right + left) / 2)
    
        temp_total = 0
        for i in range(N):
            temp_total += min(records[i], mid)
        
    
        if (temp_total > total):
            right = mid
        elif (temp_total < total):
            left = mid
            result = mid
        else:
            print(mid)
            find_flag = True
        
    if not find_flag:
        print(result)

JS代码

function main(n, records, total) {
    let total_records = eval(records.join ("+"))
    
    if (total_records <= total) 
        return -1
    
    records.sort( function (a, b){
        return a - b
    })
    
    let right = records[records.length-1]
    let left = Math.floor(total / n)
    
    let ans = left;
    
    while (right - left > 1) {
        let mid = Math.floor((right + left) / 2);
    
        let tmp = 0;
        for (let record of records) {
            tmp += Math.min(record, mid)
        }
    
        if (tmp > total) {
            right = mid;
        } else if (tmp < total) {
            left = mid;
            ans = mid;
        } else {
            console.log(mid)
            return 
        }
    }
    
    console.log(ans)
}
 
main(6,[3, 3 ,8, 7 ,10, 15],40)