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
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.
题目大意：
给出n个数据代表每天的花费，现将其任意分成m组，求满足使得每组花费最小的最大花费。
题解：
朴素穷举肯定超时，因此要使用二分法枚举答案。由题意，可行的花费介于单日花费的最大值与所有天数花费之和之间，我们再判断一下区间中值是否满足将天数分成m组的要求即可。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int n,m,a[100005];
bool check(int x)
{
int sum=0,k=1;//初始将所有天数视为一组
for(int i=0;i<n;i++)
{
sum+=a[i];
if(sum>x)
{
k++;
sum=a[i];
}
}
return k<=m;
}
int solve(int a,int b)
{
int l=a,r=b,tmp=0,mid;
while(l<=r)
{
mid=l+(r-l)/2;//此处注意下界不为0，不能写成（l+r）/2。
if(check(mid))
{
tmp=mid;
r=mid-1;
}
else
l=mid+1;
}
return tmp;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
int max=-1,sum=0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]>max) max=a[i];
sum+=a[i];
}
printf("%d
",solve(max,sum));
}
return 0;
}