1068 Find More Coins (30 分)   Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special…

1068. Find More Coins (30) 时间限制 150 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could ac…

https://pintia.cn/problem-sets/994805342720868352/problems/994805402305150976 Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805402305150976 题意: n个硬币,每一个有一个特有的价值,一个硬币只有一个,要求选取一些硬币使得他们的价值刚好是m 输出字典序最小的方案. 思路: 最近好久没有刷题了连PAT都好久没动了这样不行啊.连背包都有点不大会了.赶紧把PAT30分的刷完去刷洛谷了. 硬币是一个weight和value相同的物品,背包的容量就是m 问题转换成尽量让包中的价值大,背…

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…

//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…

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 <…

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: f…

题意: 输入两个正整数N和M(N<=10000,M<=1000),然后输入N个正整数(<=500),输出两个数字和恰好等于M的两个数(小的数字尽可能小且输出在前),如果没有输出"No Solution". AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; int main(){ ios::sync_with_stdi…

一开始没多想,虽然注意到数据N<=10^4的范围,想PAT的应该不会超时吧,就理所当然地用dfs做了,结果最后一组真的超时了.剪枝啥的还是过不了,就意识到肯定不是用dfs做了.直到看到别人说用01背包的思路,果真好久没做题了智力水平下降,且原本dp就是我的弱项,压根就没把这题往dp上去想额... (http://www.liuchuo.net/archives/2323) 题意:从n个硬皮中选取方案,使得总和价值正好为m,如果有多种,方案为排列最小的那个. 可以把硬币看成w=v(即容量=价值)的…