Coins Accepted : 120   Submit : 305 Time Limit : 1000 MS   Memory Limit : 65536 KB Coins Problem Description: Duoxida buys a bottle of MaiDong from a vending machine and the machine give her n coins back. She places them in a line randomly showing he…

题意: n个硬币摆成一排,问有连续m个正面朝上的硬币的序列种数. 很明显的DP题.定义状态dp[i][1]表示前i个硬币满足条件的序列种数.dp[i][0]表示前i个硬币不满足条件的序列种数. 那么显然有dp[i][1]=dp[i-1][1]*2+dp[i-1-m][0]. 如果前i-1个硬币满足条件,那么第i个硬币无论怎么样都满足条件.如果前i-1-m个硬币不满足条件,那么只需要再添加m个正面朝上的硬币即可. dp[i][0]=2^i-dp[i][1]. 于是最后的答案就是dp[n][1].…

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;…

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…

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…

Square Coins Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8800    Accepted Submission(s): 5991 Problem Description People in Silverland use square coins. Not only they have square shapes but…

dp[i][j]表示前i种硬币中取总价值为j时第i种硬币最多剩下多少个,-1表示无法到达该状态. a.当dp[i-1][j]>=0时,dp[i][j]=ci; b.当j-ai>=0&&dp[i-1][j-ai]>0时,dp[i][j]=dp[i-1][j-ai]-1; c.其他,dp[i][j]=-1 Source Code Problem: User: BMan Memory: 1112K Time: 1547MS Language: G++ Result: Accep…

题意: 给定一堆硬币,然后将他们分成两部分,使得两部分的差值最小;输出这个最小的差值. 思路: 想了好久都没想到一个合适的状态转移方程.后面看了别人的题解后,才知道能够转成背包问题求解. 我们将全部的硬币和的一半作为背包容量,然后将硬币的代价看成其本身的面值.然后背包中能装的最大容量 就是当中一个人分得硬币数. 代码例如以下: #include<iostream> #include<cstdio> #include<cstring> using namespace st…

https://www.codechef.com/problems/ANUCBC n个数字,选出其一个子集.求有多少子集满足其中数字之和是m的倍数.n $\le$ 100000,m $\le$ 100,最多90组数据 傻逼题模数取什么1e9+9毁我一节课该死煞笔提 [15:13:47]刚刚心塞了一会儿出去跑了几步好点了,然后发现好像是生物老师在艺术楼走廊上给人讲题(今天好像有学校给成绩好的单独上课之类的活动,好多同学都来了艺术楼的一个教室了和机房隔一个拐角........) #include <…