Dividing coins Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 558364-bit integer IO format: %lld      Java class name: Main It's commonly known that the Dutch have invented copper-wire. Two Dutch men we…

Dividing coins It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great lengt…

HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解) 题意分析 要先排序,在做01背包,否则不满足无后效性,为什么呢? 等我理解了再补上. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 505 #define nn 505*100 using namespace std;…

  Dividing coins  It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great le…

It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great length and thus crea…

It’s commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great length and thus crea…

01背包的变形. 先算出硬币面值的总和,然后此题变成求背包容量为V=sum/2时,能装的最多的硬币,然后将剩余的面值和它相减取一个绝对值就是最小的差值. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 50007 ],dp[N]; int…

link:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=503 分成2半,并且两半的差距最小,背包的体积变成V/2 #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <…

题意:给你n个硬币,和n个硬币的面值.要求尽可能地平均分配成A,B两份,使得A,B之间的差最小,输出其绝对值.思路:将n个硬币的总价值累加得到sum,   A,B其中必有一人获得的钱小于等于sum/2,另一人获得的钱大于等于sum/2.   因此用sum/2作为背包容量对n个硬币做01背包处理,   所能得到的最大容量即为其中一人获得的钱数. #include <iostream> #include <cstdio> #include <cstring> #includ…

题目描述:给出一些不同面值的硬币,每个硬币只有一个.将这些硬币分成两堆,并且两堆硬币的面值和尽可能接近. 分析:将所有能够取到的面值数标记出来,然后选择最接近sum/2的两个面值 状态表示:d[j]表示用当前给定的硬币是否可以凑得总面值j 转移方程:d[j]=d[ j-coin[i] ] 开始时只取出硬币coin[0],判断它是否能凑得总面值j 每新加入一个硬币coin[i]时,判断所有已经取出的硬币能否凑得总面值j #include <iostream> #include <cstdi…

//平分硬币问题 //对sum/2进行01背包,sum-2*dp[sum/2] #include <iostream> #include <cstring> #include <algorithm> using namespace std; ],dp[]; int main() { int n,m,sum,sum1; cin>>n; while(n--) { cin>>m; sum=; ;i<=m;i++) { cin>>val…

题目链接:https://vjudge.net/contest/103424#problem/E 题目大意: 给你一堆硬币,让你分成两堆,分别给A,B两个人,求两人得到的最小差. 解题思路: 求解两人分得钱币的最小差值,巧妙地转化为01背包问题. sum代表这堆钱币的总价值,ans=sum/2,求出得钱较少的人的钱币总量,即在这堆钱币中挑选出一定量的钱币,使得它的总值为小于或等于ans的最大值,即将它转化为01背包问题,背包容量为ans,每一个钱币看成价值与体积相等的物品. #include <…

题目链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=503 背包问题: #include <iostream> #include <string.h> #include <stdlib.h> #include <algorithm> #include <math.h> #i…

题意:一袋硬币两人分,要么公平分,要么不公平,如果能公平分,输出0,否则输出分成两半的最小差距. 思路:将提供的整袋钱的总价取一半来进行01背包,如果能分出出来,就是最佳分法.否则背包容量为一半总价的包能装下的硬币总值就是其中一个人能分得的最多的钱了,总余下的钱减去这包硬币总值.(只需要稍微考虑一下总值是奇数/偶数的问题) #include <iostream> #include <stdio.h> #include <string.h> #include <cm…

I collect and make up this pseudocode from the book: <<Introduction to the Design and Analysis of Algorithms_Second Edition>> _ Anany LevitinNote that throughout the paper, we assume that inputs to algorithms fall within their specified ranges…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

111 - History Grading LCS 103 - Stacking Boxes 最多能叠多少个box DAG最长路 10405 - Longest Common Subsequence LCS 674 - Coin Change 全然背包求方案数 10003  - Cutting Sticks 区间DP dp[l][r]代表分割l到r的最小费用 116 - Unidirectional TSP 简单递推 输出字典序最小解 从后往前推 10131 - Is Bigger Smarte…

1.hdu 1260 Tickets 题意:有k个人,售票员可以选择一个人卖,或者同时卖给相邻的两个人.问最少的售票时间. 思路:dp[i] = min(dp[i - 1] + singlep[i], dp[i - 2] + dbp[i - 1]);dp[i]表示卖到第i个人后所需最少时间.注意时间为12小时制. #include<iostream> #include<memory.h> #include<algorithm> using namespace std;…

转自:http://janfan.cn/chinese/2015/01/21/dynamic-programming.html 动态规划(Dynamic Programming,以下简称dp)是算法设计学习中的一道槛,适用范围广,但不易掌握. 笔者也是一直不能很好地掌握dp的法门,于是这个寒假我系统地按着LRJ的<算法竞赛入门经典>来学习算法,对dp有了一个比过往都更系统＼更深入的理解,并在这里写出来与大家分享. 笔者着重描述的是从穷举到dp的算法演进,并从中获取dp解法的思路,并给出多种思考…

Dividing coins Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eag…

Kingdom of Black and White Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5583 Description In the Kingdom of Black and White (KBW), there are two kinds of frogs: black frog and white frog. Now N frogs are stand…

 UVAlive 3983 Robotruck 题目: Robotruck   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description   Problem C - Robotruck Background This problem is about a robotic truck that distributes mail packages to sev…

hdu 2844 poj 1742 Coins 题目相同,但是时限不同,原本上面的多重背包我初始化为0,f[0] = 1;用位或进行优化,f[i]=1表示可以兑成i,0表示不能. 在poj上运行时间正好为时限3000ms....太慢了,hdu直接TLE(时限1s); 之 后发现其实并不是算法的问题,而是库函数的效率没有关注到.我是使用fill()按量初始化的,但是由于memset()可能是系统底层使用了四个字节拷 贝的函数(远比循环初始化快),效率要高得多..这就是为什么一直TLE的原因,fil…

You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins. Given n, find the total number of full staircase rows that can be formed. n is a non-negative integer and fits within the range of…

 Gym 101047M Removing coins in Kem Kadrãn Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Practice Description standard input/output Andréh and his friend Andréas are board-game aficionados. They know many of their friend…

UVALive - 4108 SKYLINE Time Limit: 3000MS     64bit IO Format: %lld & %llu Submit Status uDebug Description   The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would a…

UVALive - 3942 Remember the Word A potentiometer, or potmeter for short, is an electronic device with a variable electric resistance. It has two terminals and some kind of control mechanism (often a dial, a wheel or a slide) with which the resistance…

UVALive - 3942 Remember the Word Neal is very curious about combinatorial problems, and now here comes a problem about words. Know- ing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie. Since Jiejie can’t remem…

传送门 Description Hasan and Bahosain want to buy a new video game, they want to share the expenses. Hasan has a set of N coins and Bahosain has a set of M coins. The video game costs W JDs. Find the number of ways in which they can pay exactly W JDs su…

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1119 1119: Collecting Coins Time Limit: 3 Sec  Memory Limit: 128 MBSubmit: 144  Solved: 35[Submit][Status][Web Board] Description In a maze of r rows and c columns, your task is to collect as many coin…