Monthly Expense
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 7707   Accepted: 3180

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 ≤ moneyi ≤ 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


题目大意：知道农民接下来N天每天花的钱，不改变顺序要分成M份。求最大一份的最小值是多少。


二分查找，代码：














#include<iostream>  
using namespace std;  
int i,j,n,m,cxb[100010];  
bool fine(int a)  
{  
    int temp=0,x=1;  
    for(i=1;i<=n;i++)  
    {  
        if(x>m)  
            return false;  
        temp+=cxb[i];  
        if(temp>a)  
        {  
            temp=cxb[i];  
            x++;  
        }  
    }  
    if(x<=m)  
        return true;  
}  
int main()  
{  
    while(scanf("%d%d",&n,&m)!=EOF)  
    {  
        int sum=0,min=0,max,mid;  
        for(i=1;i<=n;i++)  
        {  
            scanf("%d",&cxb[i]);  
            sum+=cxb[i];  
            if(cxb[i]>min)  
                min=cxb[i]; //最小为花的最多的那天 
        }  
        max=sum;//最大为总的钱数 
        int ans; 
        while(max>=min)//进行二分查找  
        {  
            mid=(max+min)/2;  
            if(fine(mid))  
            {  
                max=mid-1;  
                ans=mid;  
            }  
            else  
            {  
                min=mid+1;  
            }  
        }  
        printf("%d\n",ans);  
    }  
    return 0;  
}