Given an array A of integer with size of n( means n books and number of pages of each book) and k people to copy the book. You must distribute the continuous id books to one people to copy. (You can give book A[1],A[2] to one people, but you cannot give book A[1], A[3] to one people, because book A[1] and A[3] is not continuous.) Each person have can copy one page per minute. Return the number of smallest minutes need to copy all the books.

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样例

Given array A = [3,2,4], k = 2.

Return 5( First person spends 5 minutes to copy book 1 and book 2 and second person spends 4 minutes to copy book 3. )

挑战

Could you do this in O(n*k) time ?

分析：只会O(n*lgM),M为所有数的总和，可以看到，求最小要多少分钟，也就是大于这个数，都是可以的，而小于这个数则都是不可以的，于是可以利用二分来找到这个点 代码：

class Solution {
public:
    /**
     * @param pages: a vector of integers
     * @param k: an integer
     * @return: an integer
     */
    int copyBooks(vector<int> &pages, int k) {
        // write your code here
        int l = 0;
        int r = 99999999;
        while(l<r)
        {
            int mid = (l+r)/2;
            if(check(mid,pages,k))
            {
                r = mid;
            }
            else
                l = mid+1;
        }
        return l;
        
    }
    bool check(int index,vector<int>& pages,int k)
    {
        int cnt = 0;
        int sum = 0;
        int i = 0;
        while(i<pages.size())
        {
            if(sum+pages[i]<=index)
            {
                sum+=pages[i++];
            }
            else if(pages[i]<=index)
            {
                cnt++;
                sum = 0;
            }
            else
                return false;
        }
        if(sum!=0)
            cnt++;
        return cnt<=k;
    }
};