Copying  Books 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=85904#problem/B 题目: Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so calledscri…

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description   Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so calledscribers. T…

题目链接: 传送门 Copying Books Time Limit: 3000MS     Memory Limit: 32768 KB Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had b…

题目连接:714 - Copying Books 题目大意:将一个个数为n的序列分割成m份,要求这m份中的每份中值(该份中的元素和)最大值最小, 输出切割方式,有多种情况输出使得越前面越小的情况. 解题思路:二分法求解f(x), f(x) < 0 说明不能满足, f(x) >= 0说明可以满足,f(x) 就是当前最大值为x的情况最少需要划分多少份-要求份数(如果f(x ) >= 0 说明符合要求而且还过于满足,即x还可以更小). 注意用long long . #include <s…

题目链接: 题目 Copying Books Time limit: 3.000 seconds 问题描述 Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after s…

  Copying Books  Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. O…

http://poj.org/problem?id=1505 Copying Books Time Limit: 3000MS   Memory Limit: 10000K Total Submissions: 7053   Accepted: 2200 Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be r…

Copying Books Time Limit: 3000MS Memory Limit: 10000K Total Submissions: 7109 Accepted: 2221 Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scri…

Copying Books 给出一个长度为m的序列\(\{a_i\}\),将其划分成k个区间,求区间和的最大值的最小值对应的方案,多种方案,则按从左到右的区间长度尽可能小(也就是从左到右区间长度构成的序列的字典序最小),\(m,k\leq 500\). 解 显然最大值的最小值想到二分,其实dp也可以,因为区间划分问题有可递推性,而且它能求出答案. 二分一个东西就等于换时间复杂度增加个\(log(n)\)增加了一个已知条件,虽然前提是单调性,但不妨以后缺少条件,先考虑二分再找单调性. 我们二分\(…

题意:把一个包含m个正整数的序列划分成k个非空的连续子序列.使得所有连续子序列的序列和Si的最大值尽量小. 二分,每次判断一下当前的值是否满足条件,然后修改区间.注意初始区间的范围,L应该为所有正整数中的最大值,否则应该判断时注意.输出解的时候要使字典序最小,所以从后面贪心. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; ll p[maxm]; bool vis[maxm]; int k,m; inli…