PAT 1068 Find More Coins[dp][难]】的更多相关文章

1068 Find More Coins (30)(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 speci…
1045 Favorite Color Stripe (30)(30 分) Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to…
1040 Longest Symmetric String (25)(25 分) Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given "Is PAT&TAP symmetric?", the longest symmetric sub-string is "s PAT&TAP s",…
满背包问题,把体积和价值看成相等的.用滚动数组优化,然后额外开辟一个choice数组来记录每次的选择,然后回溯打印.因为要按字典序,先把价值进行排序.假如选最小的商品能装满m的话,那就把判断条件改成大于等于,然后最后来 #include<bits/stdc++.h> using namespace std; ]; ]; ]; ][]; int n, m; bool cmp(int a, int b) { return a > b; } int main() { memset(flag,…
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…
HDOJ(HDU).2844 Coins (DP 多重背包+二进制优化) 题意分析 先把每种硬币按照二进制拆分好,然后做01背包即可.需要注意的是本题只需要求解可以凑出几种金钱的价格,而不需要输出种数.因此用0表示不可以,1表示可以.最后对dp数组扫描一遍即可. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 100…
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;…
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…
题意: 输入两个正整数N和M(N<=10000,M<=10000),接着输入N个正整数.输出最小的序列满足序列和为M. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; ]; ][]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n,m…