Monthly Expense（二分法）

最新推荐文章于 2020-02-24 20:06:25 发布

sdau20163940

最新推荐文章于 2020-02-24 20:06:25 发布

阅读量430

点赞数 1

CC 4.0 BY-SA版权

分类专栏：二分算法

本文链接：https://blog.youkuaiyun.com/sinat_37668729/article/details/77093507

二分算法专栏收录该内容

17 篇文章

订阅专栏

本文介绍了一个经典的预算划分问题，目标是在给定的连续日期花费数组中，将其划分为若干个连续区间，使得这些区间内的总花费最大值最小。通过使用二分查找的方法来高效地解决这一问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Monthly Expense

Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 64 Accepted Submission(s) : 19

Problem Description

Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ money_i ≤ 10,000) that he will need to spend each day over the next N (1 ≤ N ≤ 100,000) days.

FJ wants to create a budget for a sequential set of exactly M (1 ≤ M ≤ N) fiscal periods called "fajomonths". Each of these fajomonths contains a set of 1 or more consecutive days. Every day is contained in exactly one fajomonth.

FJ's goal is to arrange the fajomonths so as to minimize the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.

Input

Line 1: Two space-separated integers: N and M Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day

Output

Line 1: The smallest possible monthly limit Farmer John can afford to live with.

Sample Input

Sample Output

Source

PKU

题意：题意为给定一个n个数组成的序列，划分为m个连续的区间，每个区间所有元素相加，得到m个和，m个和里面肯定有一个最大值，我们要求这个最大值尽可能的小。

思路：用二分查找可以很好的解决这个问题。这类问题的框架为，找出下界left和上界right， while(left< right)，求出mid，看这个mid值是符合题意，继续二分。最后right即为答案。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#include <iomanip>
#define maxn 15
#define mod 1000000007
#define inf 0x3f3f3f3f
#define exp 1e-6
#define pi acos(-1.0)
using namespace std;
int a[100010];
int main()
{
    //ios::sync_with_stdio(false);
    int n,m;
    int i;
    int minn=-inf;
    int sum=0;
    cin>>n>>m;
    for(i=0;i<n;i++){ cin>>a[i]; if(minn<a[i]) minn=a[i];sum+=a[i]; }
    int left=minn,right=sum,mid;
    while(right-left>exp)
    {
        int cnt=0;
        int sum1=0;
        mid=(left+right)/2;
        for(i=0;i<n;i++)
        {
            if(sum1+a[i]>mid){cnt++;sum1=a[i];}
            else
            {
                sum1+=a[i];
            }
        }
        cnt++;   //注意此处最后一个花费的仍需要加一
        if(cnt<=m) right=mid;
        else left=mid+1;
    }
    cout<<right<<endl;
}