D - Monthly Expense

最新推荐文章于 2025-09-03 17:27:36 发布

原创最新推荐文章于 2025-09-03 17:27:36 发布 · 190 阅读

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#c++ #acm #二分法

本文探讨了一位农民如何通过合理的预算安排来最小化每月最高支出的问题。采用二分搜索算法寻找最优解，并确保预算划分符合指定的数量要求。

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

Hint

If Farmer John schedules the months so that the first two days are a month, the third and fourth are a month, and the last three are their own months, he spends at most $500 in any month. Any other method of scheduling gives a larger minimum monthly limit.

题意：

这道题就是最大值最小化，样例是说由七个数字，分成五个部分，要求的是这五个部分的最大值最小，其实最大值应该是在七个数字的最大值和它们的总和之间产生，确定了上下界之后，再进行二分，要满足两个条件，第一个就是分的组数不能不符合要求，第二个就是判断是否还有更优的情况

代码：

#include<iostream>
#include<cmath>
using namespace std;
int a[100005];
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        int sum = 0;
        int maxx = 0;
        for(int i = 0;i<n;i++)
        {
            cin>>a[i];
            sum+=a[i];
            maxx=max(maxx,a[i]);
        }
        while(maxx<sum)
        {
            int mid = (maxx+sum)/2;
            int s = 0;
            int num = 0;
            for(int i = 0;i<n;i++)
            {
                s+=a[i];
                if(s>mid)
                {
                    s=a[i];
                    num++;
                }
            }
            if(num<m)
                sum = mid;
            else maxx = mid+1;
        }
        cout<<maxx<<endl;
    }
}