题目描述
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.

给出农夫在n天中每天的花费，要求把这n天分作m组，每组的天数必然是连续的，要求分得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

输入格式
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

输出格式
Line 1: The smallest possible monthly limit Farmer John can afford to live with.

输入输出样例
输入 #1复制
7 5
100
400
300
100
500
101
400
输出 #1复制
500
说明/提示
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：有n个数，要分成m份，每份的和要尽可能小，求这个情况下和最大的一份的和。

解题思路：
二分+贪心

代码如下：

#include <iostream>
using namespace std;
typedef long long LL;
int n, k;
const int N = 100010;
int a[N];bool check(int mid) {int p = 1;LL sum = 0;for (int i = 0; i < n; i++) {if (a[i] + sum <= mid) {sum += a[i];} else {p++;sum = a[i];if (sum > mid)return false;}}return p <= k;
}int main() {LL sum = 0;cin >> n >> k;for (int i = 0; i < n; i++) {cin >> a[i];sum += a[i];}LL l = 1, r = sum;while (l < r) {LL mid = l + r >> 1;if (check(mid)) {r = mid;} else {l = mid + 1;}}cout << r << endl;return 0;
}