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…

求使最大值最小,可以想到二分答案. 然后再根据题目意思乱搞一下,按要求输出斜杠(这道题觉得就这一个地方难). Code /** * UVa * Problem#12627 * Accepted * Time:0ms */ #include<iostream> #include<cstdio> #include<cctype> #include<ctime> #include<cstring> #include<cstdlib> #in…

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: 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…

  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…

Before the invention of book-printing, it was very hard to make a copy of a book. All the contents hadto be re-written by hand by so called scribers. The scriber had been given a book and after severalmonths he finished its copy. One of the most famo…

题目描写叙述开头一大堆屁话,我还细致看了半天..事实上就最后2句管用.意思就是给出n本书然后要分成k份,每份总页数的最大值要最小.问你分配方案,假设最小值同样情况下有多种分配方案,输出前面份数小的,就像字典序输出从小到大一样的意思. 这里用到贪心的方法,定义f(x)为真的条件是满足x为最大值使n本书分成k份,那么就是求x的最小值.怎样确定这个x就是用的二分法,x一定大于0小于全部值的合,不断的二分再推断是否成立,成立就取左半边,不成立说明太小了就取右半边,写的时候还是没有把二分法理解透彻,我还怕…

题目大意: 要抄N本书,编号为1,2,3...N, 每本书有1<=x<=10000000页, 把这些书分配给K个抄写员,要求分配给某个抄写员的那些书的编号必须是连续的.每个抄写员的速度是相同的,求所有书抄完所用的最少时间的分配方案. 题目中的要求是去求划分的子序列的最大值尽量小,最大值最小化,如果从划分的角度看,无法获得好的思路,我们可以从值得角度考虑,所要求的最小的最大值必定是从[amax,sum(总和)]中取得的,那么我们可以二分法的方式猜测一个数字,看它是否满足要求,如果满足要求,我们可…

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…

题意:把一个包含m个正整数的序列划分成k个(1<=k<=m<=500)非空的连续子序列,使得每个正整数恰好属于一个序列(所有的序列不重叠,且每个正整数都要有所属序列).设第i个序列的各数之和为S(i),你的任务是让所有的S(i)的最大值尽量小.如果有多解,S(1)应尽量小,如果仍有多解,S(2)应尽量小,依此类推. 分析: 1.二分最小值x. 2.判断当前x是否满足条件时,从右往左尽量划分,若cnt<k,则从0开始依次标为分界点,这样可满足S(1),S(2),……,尽量小. #pr…