【POJ1003】Hangover(二分搜索)】的更多相关文章

[POJ1003]Hangover 试题描述 How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can m…
Hangover   Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can m…
这两题比较简单,就不做分析了,描述下题目,就上代码吧. [题目描述] 1003,其实就是求这个方程的最小n:1/2 + 1/3 + 1/4 + ... + 1/(n + 1) >= c: 1004,其实就是算个平均数,直接除12 [附:完整代码] 1003题: /* POJ-1003 Hangover */ #include <iostream> using namespace std; int main() { double c; while(cin>>c) { if (c…
直接用库函数二分即可. #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <cctype> #include <algorithm> #include <numeric> #define eps 1e-8 using namespace std; ]; int…
Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 121079   Accepted: 59223 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're a…
Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it. Note:A subtree must include all of its descendants.Here's an example: 10 / \ 15 / \ \ 1 8 7 The Largest…
Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're as…
Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7156    Accepted Submission(s): 3318 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…
Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7493    Accepted Submission(s): 3484 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,…
题目传送门 题意:问使得sum (k^i) = n || n -1 (1 <= i <= r) 的min (r*k)组合的r和k  分析:r的最大不会超过40,枚举r,二分搜索k.注意会爆long long,所以上界需要优化.r = 2开始上界就小于1e6,cyd将后面的范围也求出来了,其实1e6就够用了. 这水题卡了我好久,没有很好分析题目,做不出来就有种无力感,开始烦躁起来.还是题目做得少了,如果这种题做多了,可能看一眼就能做出来了. /**************************…
HangOver Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7693    Accepted Submission(s): 3129 Problem Description How far can you make a stack of cards overhang a table? If you have one card, yo…
Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 108765   Accepted: 53009 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're a…
还是写一下,二分搜索好了 这道题开数组比较坑... 二分,需要注意边界问题,例如:左闭右闭,左闭右开,否则查找不到or死循环 先上AC代码 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Ch{ char a[12]; char b[12]; }; int cmp(const void *aa,const…
Description Demy has n jewels. Each of her jewels has some value vi and weight wi. Since her husband John got broke after recent financial crises, Demy has decided to sell some jewels. She has decided that she would keep k best jewels for herself. Sh…
HangOver Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u   Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming…
题意:给出三种操作 0 e:将e放入容器中 1 e:将e从容器中删除,若不存在,则输出No Elment! 2 a k:搜索容器中比a大的第k个数,若不存在,则输出Not Find! 思路:树状数组+二分搜索,具体见代码吧. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> /* AC 树状数组+二分搜索 题意:给出三种操作 0 e:将e放入容器中…
一. 题目 Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 116593   Accepted: 56886 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (W…
链接地址: Poj:http://poj.org/problem?id=1003 OpenJudge:http://bailian.openjudge.cn/practice/1003 题目: Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 94993   Accepted: 46025 Description How far can you make a stack of cards overhang…
Yougth的最大化 时间限制:1000 ms | 内存限制:65535 KB 难度:4 描述 Yougth现在有n个物品的重量和价值分别是Wi和Vi,你能帮他从中选出k个物品使得单位重量的价值最大吗? 输入 有多组测试数据 每组测试数据第一行有两个数n和k,接下来一行有n个数Wi和Vi. (1<=k=n<=10000) (1<=Wi,Vi<=1000000) 输出 输出使得单位价值的最大值.(保留两位小数) 样例输入 3 2 2 2 5 3 2 1 样例输出 0.75 来源 Yo…
Description In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be . . Given your test scores and a positive integer k, determine how high you can make your cumulative aver…
Median Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5468   Accepted: 1762 Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i < j ≤ N). We can get C(N,2) difference…
Trucking Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2216    Accepted Submission(s): 757 Problem Description A certain local trucking company would like to transport some goods on a cargo…
Description Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his farm to the phone system. There are N ( ≤ N ≤ ,) forlorn teleph…
Description Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ ( ≤ i < j ≤ N). We can ) differences through this work, and now your task is to find the median of the differences as quickly as you ca…
题目简述:有一个全排列,一直每个前缀区间的逆序对数,还原这个排列. fi记录逆序对数,pi记录该位置数值,则k=fi-f(i-1)表示前i-1个数比pi大的数的个数,那么只要在剩余元素求出按大小顺序第i-k个数字即可. 线段树+二分搜索,线段树bit[i]记录i的在剩余元素的排名顺序. /******************************* Date : 2015-12-06 19:49:59 Author : WQJ (1225234825@qq.com) Link : http:/…
当我们在字典中查找某个单的时候,一般我们会翻到一个大致的位置(假设吧,翻到中间位置),开始查找.如果翻到的正好有我们要的词,那运气好,查找结束.如果我们要找的词还在这个位置的前面,那我们对前面的这一半词典继续搜索,翻到某个位置(假设是四分之一的位置)等等.这个二分搜索的工作原理一样.相应的算法就叫做二进制搜索算法. 迭代版本算法: //iterative binary search which returns index of element int binarySearchIterative(…
题目:http://poj.org/problem?id=3273 思路:通过定义一个函数bool can(int mid):=划分后最大段和小于等于mid(即划分后所有段和都小于等于mid) 这样我们转化为求 满足该函数的 最小mid.即最小化最大值,可以通过二分搜索来做,要注意二分的边界.WR了好几次. 代码: #include<iostream> #include<string> #include<cstdlib> #include<cstdio> u…
题目描写叙述 http://acm.nyist.net/JudgeOnline/problem.php?pid=306 Dr.Kong设计的机器人卡多非常爱玩.它经常偷偷跑出实验室,在某个游乐场玩之不疲. 这天卡多又跑出来了.在SJTL游乐场玩个不停,坐完碰碰车.又玩滑滑梯.这时卡多又走入一个迷宫.整个迷宫是用一个N * N的方阵给出,方阵中单元格中填充了一个整数,表示走到这个位置的难度. 这个迷宫能够向上走,向下走,向右走,向左走,可是不能穿越对角线. 走迷宫的取胜规则非常有意思,看谁能更快地…
题目连接:3971 - Assemble 题目大意:有若干个零件, 每个零件给出的信息有种类, 名称, 价格, 质量,  现在给出一个金额, 要求在这个金额范围内, 将每个种类零件都买一个, 并且尽量让这些零件中质量最小的越大, 输出质量最小的值. 解题思路:首先可以用二分搜索确定质量, 然后在搜索的过程中要判断这个质量是否能被满足, 判断函数可以用贪心, 在每一类的零件中选择价格最低且质量大于等于当前质量的零件.(事先按照价格大小排序). #include <stdio.h> #includ…
题目链接 Problem Description My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie.…