Monthly Expense解题报告

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 nextN (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 andM
Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on theith 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

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.

#include<iostream>
using namespace std;
const int MaxD=100001;
int Cost[MaxD],N,M;

int fun (const int x)  //算法实现。
{
	int s=0,m=0;
    for(int i=0;i<N;++i)
	{
		if(s+Cost[i]<=x)
			s+=Cost[i];
		else
            ++m,s=Cost[i];
	}
	return ++m;
}

int main()
{
	int max=-1,sum=0;
	scanf("%d %d",&N,&M);
	for(int i=0;i<N;++i)
	{
	    scanf("%d",&Cost[i]);
        if(Cost[i]>max)
			max=Cost[i];  //找出最大的月收入。
		sum+=Cost[i];      //算出总收入。
	}
     int left=max,right=sum,mia,k,ans;  //max<=x<=sum;x为二分查找对象。
	 while(left<=right)   //不断缩小范围。
	 {
		 mia=(left+right)/2; //中间值。
         k=fun(mia);  //返回数 和m 比较，小于m，缩小x，即right-1；否则增大x，即left+1；
		 if(k<=M)
		 {
			 right=mia-1;
		     ans=mia;
		 }
		 else
			 left=mia+1;
	 }
printf("%d\n",ans);
 return 0;
}