HDU 2955 【01背包/小数/概率DP】】的更多相关文章

Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 29499 Accepted Submission(s): 10797 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the…
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33247 Accepted Submission(s): 13449 Problem Description Speakless很早就想出国,现在他已经考完了所有需要的考试,准备了所有要准备的材料,于是,便需要去申请学校了.要申请国外的任何大学,你都要交纳一定的…
http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能超过背包的容量1-P. 在这个条件下,让装入背包的物品的总价值,也就是bag[i].[v]的和最大 bag.v是每一件物品的价值,bag.p是每件物品的体积 像上面这样想是行不通的.下面有解释 这道题麻烦的是概率这东西没法用个循环表示出来,根据我以往的经验,指望着把给出的测试数据乘上一百或者一万这种…
Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 29495 Accepted Submission(s): 10795 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the…
Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15649    Accepted Submission(s): 5747 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…
Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10933    Accepted Submission(s): 4049 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…
hdu 1203  01背包  I need a offer 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1203 题目大意:给你每个学校得到offer的概率以及花费,给定money,求得到至少一份offer的最大概率. 简单的01背包 #include<iostream> #include<cstdio> #include<cstring> using namespace std; ],p[]; int n,m; ];…
A - Robberies Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2955 Appoint description: Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usu…
I NEED A OFFER! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description Speakless很早就想出国,现在他已经考完了所有需要的考试,准备了所有要准备的材料,于是,便需要去申请学校了.要申请国外的任何大学,你都要交纳一定的申请费用,这可是很惊人的.Speakless没有多少钱,总共只攒了n万美元.他将在m个学校中选择若干的(当然要…
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40874    Accepted Submission(s): 16228   Problem Description Speakless很早就想出国,现在他已经考完了所有需要的考试,准备了所有要准备的材料,于是,便需要去申请学校了.要申请国外的任何大学,你…
http://acm.hdu.edu.cn/showproblem.php?pid=1864 New~ 欢迎“热爱编程”的高考少年——报考杭州电子科技大学计算机学院关于2015年杭电ACM暑期集训队的选拔 最大报销额 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18562    Accepted Submission(s): 5459…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1203 I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33303    Accepted Submission(s): 13470 Problem Description Speakless很早就想出国,现在…
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23039    Accepted Submission(s): 9230 Problem Description Speakless很早就想出国,现在他已经考完了所有需要的考试,准备了所有要准备的材料,于是,便需要去申请学校了.要申请国外的任何大学,你都要交…
题目链接: id=3211">poj3211  hdu1171 这个题目比1711难处理的是字符串怎样处理,所以我们要想办法,自然而然就要想到用结构体存储.所以最后将全部的衣服分组,然后将每组时间减半,看最多能装多少.最后求最大值.那么就非常愉快的转化成了一个01背包问题了... . hdu1711是说两个得到的价值要尽可能的相等.所以还是把全部的价值分为两半.最后01背包,那么这个问题就得到了解决.. 题目: Washing Clothes Time Limit: 1000MS   Me…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1171 许多有价值的物品,有重复.问如何将他们分成两堆,使两堆价值之差最小. 对价值求和,转换成01背包,做一次,相当于一堆选物品使得最接近一半.然后这个结果和用价值和作差的结果就是两堆的价值,此时价值只差最小. #include <algorithm> #include <iostream> #include <iomanip> #include <cstring&g…
http://acm.hdu.edu.cn/showproblem.php?pid=2639 01背包第k优解,把每次的max分步列出来即可 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct Node { int price; int val; }node[]; int main() { int T; scanf("%d",&…
题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1171 题意 老师有一个属性:价值(value).在学院里的老师共有n种价值,每一种价值value对应着m个老师,说明这m个老师的价值都为value.现在要将这些老师从人数上平分成两个院系,并且希望平分后两个院系老师的总价值A和B应尽可能地相等,求A和B的值(A>=B). 思路 由于每种老师的个数是有限的,所以使用多重背包解决.由于测试数据不是很严格,所以使用01背包也可以通过. 代码 01背包: #…
最大报销额 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21113    Accepted Submission(s): 6326 Problem Description 现 有一笔经费可以报销一定额度的发票.允许报销的发票类型包括买图书(A类).文具(B类).差旅(C类),要求每张发票的总额不得超过1000元,每张发 票上,单项物品…
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22869    Accepted Submission(s): 9151 Problem Description Speakless 很早就想出国,现在他已经考完了所有需要的考试,准备了所有要准备的材料,于是,便需要去申请学校了.要申请国外的任何大学,你都要…
Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3718    Accepted Submission(s): 1903 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…
Big Event in HDU Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 48364 Accepted Submission(s): 16581 Problem Description Nowadays, we all know that Computer College is the biggest department in H…
Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24321    Accepted Submission(s): 8562 Problem Description Nowadays, we all know that Computer College is the biggest department…
这题题意有点坑阿.感觉特别模糊. 我开始有一点没理解清楚.就是报销的话是整张整张支票报销的.也是我傻逼了 没一点常识 还有一点就是说单张支票总额不超过1000,每张支票中单类总额不超过600,我开始以为是题目数据会这样给. 没想到还要自己判断 并且不符合条件的支票要直接舍去.- -.只能怪自己题目做少了.   没事还是好好刷题.   纵使别人放弃 我也不允许自己放弃 这题的输入数据全部都是2位小数,可以直接化为整数处理 然后就是裸的01背包了 #include<iostream> #inclu…
转自:http://blog.csdn.net/liuqiyao_01/article/details/8753686 题意:这是又是一道01背包的变体,题目要求选出一些牛,使smartness和funness值的和最大,而这些牛有些smartness或funness的值是负的,还要求最终的smartness之和以及funness之和不能为负. 这道题的关键有两点:一是将smartness看作花费.将funness看作价值,从而转化为01背包:二是对负值的处理,引入一个shift来表 示“0”,…
饭卡 Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16453    Accepted Submission(s): 5721 Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计,即在购买之前判断余额.如果购买一个商品之前,卡上的剩余金额大于或等于5元,就一定可以购买成功(即使购买后卡上余额为负),否则无法…
ATM Mechine 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5781 Description Alice is going to take all her savings out of the ATM(Automatic Teller Machine). Alice forget how many deposit she has, and this strange ATM doesn't support query deposit. T…
果然对状压DP,我根本就不懂=.= /************************************************** Problem: 2923 User: G_lory Memory: 720K Time: 157MS Language: G++ Result: Accepted **************************************************/ #include <iostream> #include <cstring>…
http://acm.hdu.edu.cn/showproblem.php?pid=1171 基础的01背包,求出总值sum,背包体积即为sum/2 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ],bao[]; int main() { int sum,i,j,k,m,n,a; while(scanf("%d",&n)!=EOF)…
nyoj 1091 还是01背包 描述 有n个重量和价值分别为 wi 和 vi 的物品,从这些物品中挑选总重量不超过W的物品,求所有挑选方案中价值总和的最大值 1 <= n <=40 1 <= wi <= 10^15 1 <= vi <= 10^15 1 <= W <= 10^15 分析:在做的时候毫无头绪,在网上看了其他大神的博客才AC了,数据超大无法使用之前的思路(超时且dp数组开不了这么大),但是由本题条件可知n的值非常小,可用递归配合剪枝解题 代码:…
http://acm.hdu.edu.cn/showproblem.php?pid=2546 经典的01背包 预留5元买最贵的,剩余的就是01背包. #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; int main() { ],bao[],i,j,n,m,ma; while(~scanf("%d",&n),n) { memset(bao…