Gray code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 129    Accepted Submission(s): 68

Problem Description
The reflected binary code, also known as Gray code after Frank Gray, is a binary numeral system where two successive values differ in only onebit (binary digit). The reflected binary code was originally designed to prevent spurious output from electromechanical switches. Today, Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.

Now , you are given a binary number of length n including ‘0’ , ’1’ and ‘?’(? means that you can use either 0 or 1 to fill this position) and n integers(a1,a2,….,an) . A certain binary number corresponds to a gray code only. If the ith bit of this gray code is 1,you can get the point ai.
Can you tell me how many points you can get at most?

For instance, the binary number “00?0” may be “0000” or “0010”,and the corresponding gray code are “0000” or “0011”.You can choose “0000” getting nothing or “0011” getting the point a3 and a4.

 
Input
The first line of the input contains the number of test cases T.

Each test case begins with string with ‘0’,’1’ and ‘?’.

The next line contains n (1<=n<=200000) integers (n is the length of the string).

a1 a2 a3 … an (1<=ai<=1000)

 
Output
For each test case, output “Case #x: ans”, in which x is the case number counted from one,’ans’ is the points you can get at most
 
Sample Input
2
00?0
1 2 4 8
????
1 2 4 8
 
Sample Output
Case #1: 12
Case #2: 15
 
 
Hint

https://en.wikipedia.org/wiki/Gray_code

http://baike.baidu.com/view/358724.htm

 
Source
 
解题:动态规划
 
 #include <bits/stdc++.h>
using namespace std;
const int maxn = ;
int dp[maxn][],w[maxn],len;
char str[maxn];
int main() {
int kase,cs = ;
scanf("%d",&kase);
while(kase--) {
scanf("%s",str);
len = strlen(str);
for(int i = ; i < len; ++i)
scanf("%d",w + i);
memset(dp,-,sizeof dp);
if(str[] == '' || str[] == '?') dp[][] = ;
if(str[] == '' || str[] == '?') dp[][] = w[];
for(int i = ; i < len; ++i){
if(str[i] == '' || str[i] == '?'){
if(dp[i-][] != -) dp[i][] = max(dp[i][],dp[i-][]);
if(dp[i-][] != -) dp[i][] = max(dp[i][],dp[i-][] + w[i]);
}
if(str[i] == '' || str[i] == '?'){
if(dp[i-][] != -) dp[i][] = max(dp[i][],dp[i-][]);
if(dp[i-][] != -) dp[i][] = max(dp[i][],dp[i-][] + w[i]);
}
}
printf("Case #%d: %d\n",cs++,max(dp[len-][],dp[len-][]));
}
return ;
}

2015 Multi-University Training Contest 7 hdu 5375 Gray code的更多相关文章

  1. HDU 5375 Gray code (简单dp)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5375 题面: Gray code Time Limit: 2000/1000 MS (Java/Oth ...

  2. HDU 5375 Gray code(2015年多校联合 动态规划)

    题目连接 : 传送门 题意: 给定一个长度为的二进制串和一个长度为n的序列a[],我们能够依据这个二进制串得到它的Gray code. Gray code中假设第i项为1的话那么我们就能够得到a[i] ...

  3. HDU 5375 Gray code 格雷码(水题)

    题意:给一个二进制数(包含3种符号:'0'  '1'  '?'  ,问号可随意 ),要求将其转成格雷码,给一个序列a,若转成的格雷码第i位为1,则得分+a[i].求填充问号使得得分最多. 思路:如果了 ...

  4. HDU 5375 Gray code

    题意:给出一个二进制数,其中有些位的数字不确定,对于所有对应的格雷码,与一个序列a对应,第i位数字为1时得分a[i],求最大的得分. 解法:一个二进制数x对应的格雷码为x ^ (x >> ...

  5. HDU 5375——Gray code——————【dp||讨论】

    Gray code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total S ...

  6. hdu 5375 Gray code 【 dp 】

    dp[i][j]表示第i位取j的时候取得的最大的分数 然后分s[i]是不是问号,s[i-1]是不是问号这大的四种情况讨论 #include<cstdio> #include<cstr ...

  7. HDU 5375 Gray Code 动归

    题意:给你一串不确定的二进制码,其对应的格雷码的每一位有对应的权值,问转换成的格雷码的能取到的最大权值是多少. 思路:没有思路,乱搞也AC #pragma comment(linker, " ...

  8. HDU 5375 Gray code(DP)

    题意:给一串字符串,里面可能出现0,1,?,当中问号可能为0或1,将这个二进制转换为格雷码后,格雷码的每位有一个权值,当格雷码位取1时.加上该位权值,求最大权值和为多少. 分析:比赛的时候愚了.竟然以 ...

  9. hdu 5375 - Gray code(dp) 解题报告

    Gray code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total ...

随机推荐

  1. telnet允许root用户登录

    默认情况下,linux不允许root用户以telnet方式登录linux主机,若要允许root用户登录,可采取以下3种方法之一:    1.修改login文件 redhat中对于远程登录的限制体现在/ ...

  2. 移植MonkeyRunner的图片对照和获取子图功能的实现-Appium篇

    假设你的目标測试app有非常多imageview组成的话,这个时候monkeyrunner的截图比較功能就体现出来了. 而其它几个流行的框架如Robotium,UIAutomator以及Appium都 ...

  3. hdu2688 Rotate(树状数组)

    题目链接:pid=2688">点击打开链接 题意描写叙述:对一个长度为2<=n<=3000000的数组,求数组中有序对(i<j而且F[i]<F[j])的数量?其 ...

  4. 原来C++之父在大摩工作呀,并且还是总经理。。

    摩根士丹利信息技术部门简历接收即将截止.请同学们抓紧投递 摩根士丹利9月.10月将在中国各大高校举办包含技术讲座.信息分享会以及校园宣讲会在 内的一系列校园活动.同学们将有机会和摩根士丹利高管以及返校 ...

  5. iOS-UIImageView载入网络下载的图片(异步+多线程)

    最原始的载入网络下载的图片方式: //最原始载入网络图片方法,相当堵塞主线程,界面卡顿 -(void)setImageWithURL:(NSString *)imageDownloadUrl{ UII ...

  6. QFileSystemModel只显示名称,不显示size,type,modified

    Qt 提供的 QFileSystemModel可以提供文件目录树预览功能,但是预览的都自带了Name,size,type, modified等信息.我现在只想显示name这一列,不想显示size,ty ...

  7. Android获取系统时间的多种方法

    Android中获取系统时间有多种方法,可分为Java中Calendar类获取,java.util.date类实现,还有android中Time实现. 现总结如下: 方法一: ? 1 2 3 4 5 ...

  8. java中字符串编码转换

    Java 正确的做字符串编码转换 字符串的内部表示? 字符串在java中统一用unicode表示( 即utf-16 LE) , 对于 String s = "你好哦!"; 如果源码 ...

  9. maven 项目加载本地JAR

     将jar安装到本地的maven仓库 1.首先确定本地有maven环境. 2.安装本地jar 模板: mvn install:install-file -Dfile=<path-to-file& ...

  10. Mac 安装cmake小问题

    今天用 brew install cmake. ==> Downloading https://homebrew.bintray.com/bottles/cmake-3.9.6.sierra.b ...