题意:有N件湿的衣服,一台烘干机.每件衣服有一个湿度值.每秒会减一,如果用烘干机,每秒会减k.问最少多久可以晒完. 题解:二分.首先时间越长越容易晒完. 其次判定函数可以这样给出:对于答案 X,每一个湿度大于X的衣服都要被烘干.所以可以直接统计烘干机被用的总时间,如果其大于X则返回0. 注意向下取整细节. #define _CRT_SECURE_NO_WARNINGS #include<cstring> #include<cctype> #include<cstdlib>…

Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12568   Accepted: 3243 Description It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has…

It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has decided to use a radiator to make drying faster. But the radiator is small, so it can hold only one thi…

烘干衣服 题目大意:主人公有一个烘干机,但是一次只能烘干一件衣服,每分钟失水k个单位的水量,自然烘干每分钟失水1个单位的水量(在烘干机不算自然烘干的那一个单位的水量),问你最少需要多长时间烘干衣服? 简单来说题目就是要:时间允许的情况下让时间最小,时间可以无限大,这题就是“最小化最大值”,二分法 #include <iostream> #include <functional> #include <algorithm> using namespace std; ]; v…

传送门 表示又是神题一道 Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9327   Accepted: 2364 Description It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring proces…

依然是学习分析方法的一道题 求一个长度为n的序列中的一个平均值最大且长度不小于L的子段,输出最大平均值 最值问题可二分,从而转变为判定性问题:是否存在长度大于等于L且平均值大于等于mid的字段和 每个数与mid作差再转变为求非负子段 子段和问题应该利用前缀和C,长度大于等于L的字段和最大值可表示为 max{Aj+1 + Aj+2 ... + Ai},i-j+1-1>=L,j+1>=1 等价于 max{Ci-min{Cj}},L<=i<=n,0<=j<=i-L 注意是i=…

题目链接:http://poj.org/problem?id=3104                                                                                                      Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11128   Accepted: 2865 Description It is very…

http://poj.org/problem?id=3104 题目大意: 有n件衣服,每件有ai的水,自然风干每分钟少1,而烘干每分钟少k.求所有弄干的最短时间. 思路: 注意烘干时候没有自然风干. 能够理解为烘干时每分钟掉(k-1)的水. 这样每件衣服每分钟就都掉1水了. 二分枚举最小值就可以. #include<cstdio> #include<algorithm> using namespace std; const int MAXN=100000+10; int a[MAX…

[题目链接] http://poj.org/problem?id=3104 [题目大意] 给出n件需要干燥的衣服,烘干机能够每秒干燥k水分, 不在烘干的衣服本身每秒能干燥1水分 求出最少需要干燥的时间. [题解] 考虑将烘干机的烘干效应变为k-1,那么就是每件衣服在每秒都会自动减少一水分 如果我们知道最少需要的时间,那么每件衣服自己减少的水分数量就知道了, 在除去自然减少的水分之后,看看还需要多少k-1的水分减少才能烘干全部的衣服就可以了, 因此我们二分这个答案,验证是否可行即可. [代码] #…

题目链接:http://poj.org/problem?id=3104 Drying Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9440   Accepted: 2407 Description It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afrai…