二分算法（最大值最小化）

Monthly Expense

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.

给定一个大小为n的数组，假设分成m个子序列，找出m个序列中的最大值的最小值。例如序列： 1 2 3 2 5 4 划分成3个序列的最有方案为： 1 2 3 | 2 5 | 4，其中S(1)、S(2)、S(3) 分别为 6 、7、 4，最大值为 7；如果划分成 1 2 | 3 2 | 5 4，则最大值为 9。则答案为7。

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 100005;
int n, m;
int a[maxn];
bool isPart(int x)
{
	int sum = 0;
	int line = 0;
	for (int i = 1;i <= n;i++)
	{
		if (sum + a[i] < x)
		{
			sum += a[i];
		}
		else
		{
			sum = a[i];
			line++;
			if (line > m - 1)
			{
				return false;
			}
		}
	}
	return true;
}
int main()
{
	while (cin >> n >> m && n&&m)
	{
		double low = 0.0;
		double high = 0.0;
		for (int i = 1;i <= n;i++)
		{
			cin >> a[i];
			high += a[i];
			if (low < a[i])
			{
				low = a[i];
			}
		}
		int mid = (high + low) / 2;
		while (low < high)
		{
			if (isPart(mid))
			{
				high = mid - 1;
			}
			else
			{
				low = mid + 1;
			}
			mid = (low + high) / 2;
		}
		cout << mid << endl;
	}
	return 0;
}