Time Limit: 1000MS   Memory Limit: 10000KB   64bit IO Format: %lld & %llu Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N di…

题目链接:https://cn.vjudge.net/problem/POJ-1276 题意 懒得写了自己去看好了,困了赶紧写完这个回宿舍睡觉,明早还要考试. 思路 多重背包的二进制优化. 思路是将n个物品拆分成log(m)个物品,可使得这些物品组合出1~n个原物品,这个用于01背包中. 提交过程 WA 没理解num-=k AC 代码 #include <cstdio> #include <cstring> #include <algorithm> using name…

题目大意 给定一个容量为M的背包以及n种物品,每种物品有一个体积和数量,要求你用这些物品尽量的装满背包 题解 就是多重背包~~~~用二进制优化了一下,就是把每种物品的数量cnt拆成由几个数组成,1,2,4,~~~cnt-2^K+1,k满足cnt-2^K+1>0的最大整数,体积和价值乘上相应的数就是相应物品的价值和体积,这样用这些物品能够表示1~~cnt所有的情况~~~这就转化成01背包了~~~ 代码: #include <iostream> #include <cstdio>…

题目:http://poj.org/problem?id=1276 多重背包模板题,没什么好说的,但是必须利用二进制的思想来求,否则会超时,二进制的思想在之前的博客了有介绍,在这里就不多说了. #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> #include <math.h> using namespace std; ],cnt[],d…

Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26172   Accepted: 9238 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The ma…

Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33444   Accepted: 12106 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The m…

Cash Machine Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 24213 Accepted: 8476 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machin…

<span style="color:#000099;">/* Cash Machine Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 26604 Accepted: 9397 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bi…

个人心得:多重背包,自己根据转换方程写总是TLE,后面去网上看了二进制转换,不太理解: 后面仔细想了下,用自己的思想理解下把,就是将对应number,cash总和用二进制拆分, 然后全部装入到一个数组,这样子就可以减少循环,同时转变为01背包,这样子想把, 5 5,就变成了5,10,20,5然后用01背包互相取与不取也可以组成对应的k=1-5乘以5的值. 转换方式如下: ) { t[account++]=k*cash[i]; number[i]-=k; k=k*; } t[account++]=…

Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 38986   Accepted: 14186 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The m…

Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 30006   Accepted: 10811 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The m…

Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the…

最近的很多题解应该都是dp相关的了,emmm因为dp对我而言思考难度比较大,那么为了理顺自己的思路当然只能通过写blog整理了.愿我能成功搞定dp这个大关!(至少中等难度的dp要能够解决啊o(TヘTo)) 题意与分析 这条题目是一条明显的多重背包题目.遇到这种题目,我们首先简单的转化为01背包问题来解决.(思路:把$n_i$个$v_i$拆分开来)但是呢,有一个二进制优化可以采用,就是在这个拆分机制上.重点就是这段代码: while(num-k>=0) { things[cnt++]=d*k; n…

大神博客转载http://www.cppblog.com/MatoNo1/archive/2011/07/05/150231.aspx多重背包的单调队列初中就知道了但一直没(不会)写二进制优化初中就写过一直不写会心虚就写一下这个吧朴素方程dp[i,j]=max(dp[i-1,j-w[i]*k]+c[i]*k) w[i]*k<=j k<=j div w[i]忽略第一维dp[j]=max(dp[j-w[i]*k]+c[i]*k)复杂度O(m2n)以下是大神原文:将决策下标按照模w0的余数进行分类,…

题意: 有n种纸币,已知每种纸币的面值和数量,求所能凑成的不超过cash的最大总面值. 分析: 这道题自己写了一下TLE了,好可耻.. 找了份比较简洁的代码抄过来了..poj1276 #include <cstdio> #include <cstring> ; + ; bool vis[maxp];//是否到达总面值i int used[maxp];//到达总面值i时,该种纸币所用的数量 int a[maxn], b[maxn];//每种纸币的数量以及面值 int cash, n;…

题目地址:http://poj.org/problem?id=1276 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24067   Accepted: 8414 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses ex…

Problem Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denominatio…

Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 35387   Accepted: 12816 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses e…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44409   Accepted: 16184 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses e…

http://poj.org/problem?id=1276 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #define For(n) for(int i=1; i<=(n); i++) #define maxn 100010 using namespace std; int dp[maxn]; struct node { int n; int m; }p…

转载地址:http://m.blog.csdn.net/blog/u010489766/9229011 题目链接:http://poj.org/problem?id=1276 题意:机器里面共有n种面额的钱币,每种各ni张,求机器吐出小于等于所要求钱币的最大值 解析1:这题大牛都用了多重背包,不过,我一同学想出了一种就这题而言特别简单有效的方 法.(话说我就认为这题本来就不需要用到背包的,因为n的范围只到10,太小了).方法就是对钱进行遍历,看这些钱一共能组成多少面额的钱,然后从 cash向下枚…

多重背包的模型,但一开始直接将N个物品一个一个拆,拆成01背包竟然T了!!好吧OI过后多久没看过背包问题了,翻出背包九讲看下才发现还有二进制优化一说........就是将n个物品拆成系数:1,2,4,8....*物品价值和空间的物品,在这题中只要乘上money[i]就行了,从二进制考虑发现,这样可以组成0~n中所有的数 #include <iostream> #include <cstdio> #include<string.h> using namespace std…

http://poj.org/problem?id=1276 #include <stdio.h> #include <string.h> ; #define Max(a,b) (a)>(b)?(a):(b) int dp[N],cash; void ZeroOnePack(int cost)//01背包 { for (int i = cash; i >= cost; i--) { dp[i] = Max(dp[i],dp[i-cost]+cost); } } void…

题目网址:http://poj.org/problem?id=1276 思路: 很明显是多重背包,把总金额看作是背包的容量. 刚开始是想把单个金额当做一个物品,用三层循环来 转换成01背包来做.T了…… 后面学习了 用二进制来处理数据. 简单地介绍一下二进制优化:✧(≖ ◡ ≖✿)  假设数量是8,则可以把它看成是1,2,4,1的组合,即这4个数的组合包括了1-8的所有取值情况.这是为什么呢?将它们转换成二进制再观察一下: 1:1 2:10 4:100 1:1 二进制都只有0,1.所以1,2,4…

01背包加变形 动态规划的时候就犯浑了,每个状态都要记录的,我却只记录了当前状态的!! #include<stdio.h> #include<string.h> int max(int a,int b) { return (a) > (b) ? (a) : (b); } ],b[],M,dp[][]; int main(){ int N,i,j,k,ma; while(scanf("%d",&M)!=EOF) { scanf("%d&qu…

题意:多重背包模型  n种物品 每个m个  问背包容量下最多拿多少 这里要用二进制优化不然会超时 #include<iostream> #include<cstdio> #include<cstring> using namespace std; +; int dp[maxn]; ],c[]; int main(){ int n,m,maxnum; while(cin>>maxnum>>n){ int a,b; ; ;i<=n;i++){…

解法 多重背包板子题 多重背包板子 如果上限的体积大于了给定的体积那么套完全背包 否则二进制优化成01背包 代码 #include <iostream> #include <cstring> using namespace std; int dp[100005],m; void zb(int v,int w) { for(int i=m;i>=v;i--) dp[i]=max(dp[i],dp[i-v]+w); } void cb(int v,int w) { for(int…

链接:POJ - 1276 题意:给你一个最大金额m,现在有n种类型的纸票,这些纸票的个数各不相同,问能够用这些纸票再不超过m的前提下凑成最大的金额是多少? 题解:写了01背包直接暴力,结果T了,时间复杂度太高了,要跑外循环m和内循环所有的纸票的个数.这个题需要把每种纸票的的个数存的时候转化成2的次幂的形式来存,比如有8个$1,就可以存成1,2,4,1.这样就可以不存放8个1了,如果在个数大的情况下存储的也不会很多,因为转化成这样子,在1~8每一个都可以凑出来,随机搭配,另一方面,如果是8个$2…