Distribution of books
Time Limit: 8000/8000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
zz6d likes reading very much, so he bought a lot of books. One day, zz6d brought n books to a classroom in school. The books of zz6d is so popular that K students in the classroom want to borrow his books to read. Every book of zz6d has a number i (1<=i<=n). Every student in the classroom wants to get a continuous number books. Every book has a pleasure value, which can be 0 or even negative (causing discomfort). Now zz6d needs to distribute these books to K students. The pleasure value of each student is defined as the sum of the pleasure values of all the books he obtains.Zz6d didn't want his classmates to be too happy, so he wanted to minimize the maximum pleasure of the K classmates. zz6d can hide some last numbered books and not distribute them,which means he can just split the first x books into k parts and ignore the rest books, every part is consecutive and no two parts intersect with each other.However,every classmate must get at least one book.Now he wonders how small can the maximum pleasure of the K classmates be.
Input contains multiple test cases.
For each case, print the smallest maximum pleasure of the K classmates, and one line one case.
2 4 2 3 -2 4 -2 5 4 -1 -1 -1 -1 6
In the first example,classmate 1 get book 1,2, classmate 2 get book 3,4.the maximum pleasure is max((3-2),(4-2))=2; In the second example,he can ignore book 5 and spilt the first 4 books into 4 parts,give them to his classmates.
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将一个长度为n的数组a、a、… 、a[n]， 要求取前k段（要求各段连续，但不可交叉，每段元素个数≥1），使得最大的那段和最小，并输出该和。
1 <= n <= 2*105
1 <= k <= n
-109 <= ai <= 109