Monthly Expense

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

   
     

      
    7 5
100
400
300
100
500
101
400
   
     

 

Sample Output

   
     

      
    500
   
     

 

#include<iostream>
#include<cstdio>
using namespace std;
int n,m;
int a[100010];
int fun(int s)   //fun函数是推断当前的mid值能把n分成几组
{                //通过比較q与m的大小，对mid的值进行优化
    int sum=0,q=1;
    for(int i=1;i<=n;i++)//从第一天開始向下遍历每天的花费
    {
        if(sum+a[i]<=s)//前第I天和<=mid时，把他们归为这一组
            sum+=a[i];
        else//否则。第i天做为下一组的第一天
        {
            sum=a[i];
            q++;//组数加一
        }
    }
    if(q>m)//组数大于m，代表mid数值小了，
        return 0;
    return 1;//否则大了
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        int l,r=0,mid,min=0,sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
            if(a[i]>r)r=a[i];
        }
        l=sum;
        int ans=0;
        mid=(l+r)/2;
        while(r<l)
        {
            if(!fun(mid))//假设mid偏小
            {

               r=mid+1;//下界上升
            }
            else//否则
            {
                l=mid-1;//上限下降
            }
             mid=(r+l)/2;
        }
        cout<<mid<<endl;
    }
    return 0;
}