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
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：
给出农夫在n天中每天的花费，要求把这n天分成m组，每组的天数必须连续的，要求各组的花费之和应该尽可能地小，最后输出各组花费之和中的最大值。
就是用二分来做，二分找最大值来划分组，如果可以划分为m个组即可以找到各组中的最大值。
代码：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int n,m; //n天 m组
int a[100007];
int HF(int mid)
{
int ans=0; //组数
int sum=0; //各个组的和
for(int i=0;i<n;i++)
{
if(sum+a[i]<=mid) //如果这个组的和≤mid，那么就还可以加入这一天的花费
{
sum+=a[i];
}
else //否则就将这一天划到下一组中
{
sum=a[i];
ans++;
}
}
ans++; //最后一份也是一组
if(ans<=m)return 1; //如果组数≤m，说明mid找大了
else return 0;
}
int main()
{
scanf("%d%d",&n,&m);
int maxn=0; //花费最高的一天
int sum=0; //n天的总费用
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
maxn=max(maxn,a[i]);
sum+=a[i];
}
int l=maxn,r=sum;
int mid;
while(l<=r)
{
mid=(l+r)/2;
if(HF(mid)) //通过mid划分的组数≤m，mid划分大了
{
r=mid-1;
}
else
{
l=mid+1;
}
}
printf("%d\n",mid);
return 0;
}