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…

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…

  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…

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 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…

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…

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…