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): 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; ];…

Rikka with Subset Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1122    Accepted Submission(s): 541 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation…

dp[i][j]表示前i个元素,子集和为j的个数.d[i][j] = d[i][j] + d[i-1][j-k] (第i个元素的值为k).这里可以优化成一维数组 比如序列为 1 2 3,每一步的dp值为 1 0 0 0 0 0 0 (d[0][0]=1) 1 1 0 0 0 0 0 1 1 1 1 0 0 0 1 1 1 2 1 1 1 最终的序列就是题目给出的B序列,把B序列减去每一次dp得到的序列,第一个非0值就是a序列中的值 #include <cstdio> #include <…

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…

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…

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…

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…