leetcode 1231.Divide Chocolate (数组分段和最小值的最大值）

比较简单，就不翻译了
1231. Divide Chocolate

You have one chocolate bar that consists of some chunks. Each chunk has its own sweetness given by the array sweetness.

You want to share the chocolate with your K friends so you start cutting the chocolate bar into K+1 pieces using K cuts, each piece consists of some consecutive chunks.

Being generous, you will eat the piece with the minimum total sweetness and give the other pieces to your friends.

Find the maximum total sweetness of the piece you can get by cutting the chocolate bar optimally.

Example 1:

Input: sweetness = [1,2,3,4,5,6,7,8,9], K = 5
Output: 6
Explanation: You can divide the chocolate to [1,2,3], [4,5], [6], [7], [8], [9]

Example 2:

Input: sweetness = [5,6,7,8,9,1,2,3,4], K = 8
Output: 1
Explanation: There is only one way to cut the bar into 9 pieces.

Example 3:

Input: sweetness = [1,2,2,1,2,2,1,2,2], K = 2
Output: 5
Explanation: You can divide the chocolate to [1,2,2], [1,2,2], [1,2,2]

Constraints:

0 <= K < sweetness.length <= 10^4
1 <= sweetness[i] <= 10^5

一开始想了半天，搞了个O(n^2*k) DP，理所当然的TLE。苦思冥想之下打开百度作弊 查找资料，发现了分段和最大值最小问题，然后发现自己是个傻逼 大概是二分查找极不熟悉的后果。

class Solution {
public:
    int divis(vector<int>& s,int sum){//满足最小分段和为sum情况下的分段上限
        int now=0,r=0;
        for(int i=0;i<s.size();++i){
            now+=s[i];
            if(now>=sum){
                ++r;
                now=0;
            }
        }
        return r;
    }
    int maximizeSweetness(vector<int>& s, int k) {
        int t=(1<<30),r=0;
        while(t>0){
            if(divis(s,r+t)>k) r+=t;
            t>>=1;
        }
        return r;
    }
};

Time O(nlog(sum(s)))
Memory O(1) (extra)