CRB and His Birthday Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 357 Accepted Submission(s): 191 Problem Description Today is CRB's birthday. His mom decided to buy many presents for her lovel…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5410 Problem Description Today is CRB's birthday. His mom decided to buy many presents for her lovely son.She went to the nearest shop with M Won(currency unit).At the shop, there are N kinds of pr…

CRB and His Birthday Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 430    Accepted Submission(s): 237 Problem Description Today is CRB's birthday. His mom decided to buy many presents for her…

Problem Description Today is CRB's birthday. His mom decided to buy many presents for her lovely son. She went to the nearest shop with M Won(currency unit). At the shop, there are N kinds of presents. It costs Wi Won to buy one present of i-th kind.…

对于每个物品,如果购买,价值为A[i]*x+B[i]的背包问题. 先写了一发是WA的= =.代码如下: #include <stdio.h> #include <algorithm> #include <string.h> #include <set> using namespace std; typedef pair<int,int> pii; pii dp[]; ],A[],B[]; int main() { int T;scanf(&quo…

传送门:点击打开链接 题意:你有M块钱,如今有N件商品 第i件商品要Wi块,假设你购买x个这种商品.你将得到Ai*x+Bi个糖果 问能得到的最多的糖果数 思路:很好的一道01背包和全然背包结合的题目 首先,对于第i件商品,假设仅仅买1个,得到的价值是Ai+Bi 假设在买1个的基础上再买.得到的价值就是Ai 也就是说,除了第一次是Ai+Bi.以后购买都是Ai 那么,我们是否能将i商品拆分成两种商品,当中两种商品的代价都是Wi, 第一种的价值是Ai+Bi,可是仅仅同意买一次 另外一种的价值是Ai.能…

题意:有n种商品,每种商品中有a个糖果,如果买这种商品就送多b个糖果,只有第一次买的时候才送.现在有m元,最多能买多少糖果? 思路:第一次买一种商品时有送糖果,对这一次进行一次01背包,也就是只能买一次.然后对这种商品来一次完全背包,此时不送糖果,也可以多买. #include <bits/stdc++.h> #define pii pair<int,int> #define INF 0x7f7f7f7f #define LL long long using namespace s…

题目传送门 题意:有n个商店,有m金钱,一个商店买x件商品需要x*w[i]的金钱,得到a[i] * x + b[i]件商品(x > 0),问最多能买到多少件商品 01背包+完全背包:首先x == 1时,得到a[i] + b[i],若再买得到的是a[i],那么x == 1的情况用01背包思想,x > 1时就是在01的基础上的完全背包.背包dp没刷过专题,这么简单的题也做不出来:( ;;;; ; ;}…

#include<stdio.h> #include<string.h> #include<vector> #include<queue> #include<algorithm> using namespace std; int main() { int _,i,j,m,n,k,a[1024],b[1024],w[1024],dp[2048]; scanf("%d\n",&_); while(_--) { scanf(…

题目地址:HDU 5410 题意:有M元钱,N种礼物,若第i种礼物买x件的话.会有Ai*x+Bi颗糖果,现给出每种礼物的单位价格.Ai值与Bi值.问最多能拿到多少颗糖果. 思路:全然背包问题. dp[j][1]在当前物品时花钱为j的而且买过当前物品的最大值. dp[j][0]不买当前这件物品此前花钱为j的的最大值. 每种物品的价值随Ai线性变化,可是不随B[i]线性变化.B[i]仅是在第一次挑选第i件物品是才算入,其它时候均不算入. 所以我们能够写出状态转移方程: dp[j][0]=max(dp…