G - Gray Code
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87643#problem/G

Description

Denis, Vanya and Fedya gathered at their first team training. Fedya told them that he knew the algorithm for constructing a Gray code.
  1. Create a 2-bit list: {0, 1}.
  2. Reflect this list and concatenate it with the original list: {0, 1, 1, 0}.
  3. Prefix old entries with 0, and prefix new entries with 1: {00, 01, 11, 10}.
  4. Repeat steps 2 and 3 until the length of all elements is equal to n.

The number n is a length of a Gray code. For example, the code of length 3 is: {000, 001, 011, 010, 110, 111, 101, 100}.

Denis ran the Fedya's algorithm and obtained a binary number x at position k (positions are numbered starting from zero). Vanya wrote down the numbers k and x in binary system. This story happened many years ago and now you hold the paper sheet with these numbers in your hands. Unfortunately, some digits are unreadable now. Could you determine the values of these digits using the readable digits?

Input

The first line contains a number k written in the binary system. Unreadable digits are denoted with symbol “?”. The second line contains a number x in the same format. The lengths of these numbers are equal and don't exceed 10 5. The numbers may contain leading zeroes.

Output

If there is a unique way to restore the numbers k and x, output them, replacing the symbols “?” with “0” or “1”. If there are multiple ways to restore them, output “Ambiguity”. If Denis or Vanya certainly made a mistake in these numbers, output “Impossible”.

Sample Input

0?1
0?0

Sample Output

011
010

HINT

题意

题目给了一种序列的构造方式

然后给你一个k,判断第k个数是不是说给的x

题解

找规律题,注意翻转= =

代码:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
#include <bitset>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1) const char * fail = "Impossible";
const char * more = "Ambiguity";
using namespace std;
const int maxn = 1e5 + ;
int dp[maxn][];
char s1[maxn] , s2[maxn]; int main(int argc,char *argv[])
{
scanf("%s%s",s1+,s2+);
int Length = strlen(s1+);
s1[] = '';
for(int i = Length ; i > ; -- i)
{
if (s2[i] == '')
{
//同号
if (s1[i] == '' && s1[i-] == '') continue;
if (s1[i] == '' && s1[i-] == '')
{
printf("%s\n",fail);
return ;
}
if(s1[i] == '' && s1[i-] == '?')
{
s1[i-] = '';
continue;
}
if (s1[i] == '' && s1[i-] == '')
{
printf("%s\n",fail);
return ;
}
if (s1[i] == '' && s1[i-] == '') continue;
if (s1[i] == '' && s1[i-] == '?')
{
s1[i-] = '';
continue;
}
if (s1[i] == '?' && s1[i-] == '')
{
s1[i] = '';
continue;
}
if (s1[i] == '?' && s1[i-] == '')
{
s1[i] = '';
continue;
}
if (s1[i] == '?' && s1[i-] == '?') continue;
}
if (s2[i] == '')
{
//异号
if (s1[i] == '' && s1[i-] == '') continue;
if (s1[i] == '' && s1[i-] == '')
{
printf("%s\n",fail);
return ;
}
if(s1[i] == '' && s1[i-] == '?')
{
s1[i-] = '';
continue;
}
if (s1[i] == '' && s1[i-] == '') continue;
if (s1[i] == '' && s1[i-] == '')
{
printf("%s\n",fail);
return ;
}
if(s1[i] == '' && s1[i-] == '?')
{
s1[i-] = '';
continue;
}
if(s1[i] == '?' && s1[i-] == '')
{
s1[i] = '';
continue;
}
if(s1[i] == '?' && s1[i-] == '')
{
s1[i] = '';
continue;
}
if(s1[i] == '?' && s1[i-] == '?') continue;
}
}
for(int i=;i<=Length;i++)
if(s1[i]=='?')
{
if(s2[i]!='?')
{
if(s2[i]=='') s1[i]=s1[i-];else s1[i]=''+-(s1[i-]-'');
continue;
}
printf("%s\n",more);
return ;
}
for(int i=;i<Length;i++)
if(s1[i]!=s1[i+])
{
s2[i+]='';
}
else s2[i+]='';
printf("%s\n%s\n",s1+,s2+);
return ;
}

URAL 1780 G - Gray Code 找规律的更多相关文章

  1. BZOJ 1228 E&G(sg函数+找规律)

    把一对石子堆看出一个子游戏.打出子游戏的sg表找规律.. 这个规律我是一定找不出来的... 对于i,j,如果 (i-1)%pow(2,k+1) < pow(2,k) (j-1)%pow(2,k+ ...

  2. 牛客小白月赛 G 异或 找规律

    链接:https://www.nowcoder.com/acm/contest/135/G来源:牛客网 题目描述 从前,Apojacsleam家的水族箱里,养了一群热带鱼. 在这几条热带鱼里,Apoj ...

  3. POJ 1850 Code(找规律)

    Code Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7913   Accepted: 3709 Description ...

  4. 2018南京区域赛G题 Pyramid——找规律&&递推

    先手动推出前10项,再上BM板子求出递推式 $A_n = 5A_{n-1} - 10A_{n-2} + 10A_{n-3} - 5A_{n-4} + A_{n-5}$,根据特征根理论可求出特征方程 $ ...

  5. Ural 1780 Gray Code 乱搞暴力

    原题链接:http://acm.timus.ru/problem.aspx?space=1&num=1780 1780. Gray Code Time limit: 0.5 secondMem ...

  6. 牛客小白月赛5 G 异或(xor) 【找规律】

    题目链接: https://www.nowcoder.com/acm/contest/135/g 题目描述 从前,Apojacsleam家的水族箱里,养了一群热带鱼. 在这几条热带鱼里,Apojacs ...

  7. ACM-ICPC 2018 焦作赛区网络预赛 G. Give Candies (打表找规律+快速幂)

    题目链接:https://nanti.jisuanke.com/t/31716 题目大意:有n个孩子和n个糖果,现在让n个孩子排成一列,一个一个发糖果,每个孩子随机挑选x个糖果给他,x>=1,直 ...

  8. Ural 2045. Richness of words 打表找规律

    2045. Richness of words 题目连接: http://acm.timus.ru/problem.aspx?space=1&num=2045 Description For ...

  9. Ural 2037. Richness of binary words 打表找规律 构造

    2037. Richness of binary words 题目连接: http://acm.timus.ru/problem.aspx?space=1&num=2037 Descripti ...

随机推荐

  1. 【转】Android Studio -修改LogCat的颜色*美爆了*

    原文网址:http://www.2cto.com/kf/201505/400357.html 一. 先看效果 二.设置 File->Settings 或Ctrl + Alt +S 找到 Edit ...

  2. TCP/IP详解学习笔记(11)-TCP交互数据流,成块数据流

    目前建立在TCP协议上的网络协议特别多,有telnet,ssh,有ftp,有http等等.这些协议又可以根据数据吞吐量来大致分成两大类:(1)交互数据类型,例如telnet,ssh,这种类型的协议在大 ...

  3. Ejabberd源码解析前奏--配置

    一.基本配置     配置文件将在你第一次启动ejabberd时加载,从该文件中获得的内容将被解析并存储到内部的ejabberd数据库中,以后的配置将从数据库加载,并且任何配置文件里的命令都会被添加到 ...

  4. [转] arcgis Engine创建shp图层

    小生 原文 arcgis Engine创建shp图层 以创建点图层为例.首先要得到保存文件的地址. SaveFileDialog saveFileDialog = new SaveFileDialog ...

  5. WebForm+Web.config: 超时时间已到。在操作完成之前超时时间已过或服务器未响应。

    ylbtech-Error-WebForm+Web.config: 超时时间已到.在操作完成之前超时时间已过或服务器未响应. 超时时间已到.在操作完成之前超时时间已过或服务器未响应. 1.A,错误代码 ...

  6. HDU 3695-Computer Virus on Planet Pandora(ac自动机)

    题意: 给一个母串和多个模式串,求模式串在母串后翻转后的母串出现次数的的总和. 分析: 模板题 /*#include <cstdio> #include <cstring> # ...

  7. Linux Systemd——在RHEL/CentOS 7中启动/停止/重启服务

    RHEL/CentOS 7.0中一个最主要的改变,就是切换到了systemd.它用于替代红帽企业版Linux前任版本中的SysV和Upstart,对系统和服务进行管理.systemd兼容SysV和Li ...

  8. kmeans算法的matlab实践

    把图像中所有的像素点进行RGB聚类分析,然后输出看结果 img = imread('qq.png'); %取出R矩阵,并将这个R矩阵拉成一列 imgR = img(:,:,1); imgR = img ...

  9. SSH无法连接服务器

    服务器版本如下: @kelWEB4:/etc# lsb_release -a LSB Version: :core-4.0-amd64:core-4.0-noarch:graphics-4.0-amd ...

  10. LeetCode题解——3Sum

    题目: 给定一个数组,找出其中和为0的所有3个数的组合.每个组合的3个数都是非递降的. 解法: 先排序再遍历,设置3个指针,第一个依次遍历,第二三个在第一个指针后面的部分里,左右夹逼查找和为第一个数的 ...