Generalized Palindromic Number


Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                            

A number that will be the same when it is written forwards or backwards is known as a palindromic number. For example, 1234321 is a palindromic number.

We call a number generalized palindromic number, if after merging all the consecutive same digits, the resulting number is a palindromic number. For example, 122111 is a generalized palindromic number. Because after merging, 122111 turns into 121 which is a palindromic number.

Now you are given a positive integer N, please find the largest generalized palindromic number less than N.

Input

There are multiple test cases. The first line of input contains an integer T (about 5000) indicating the number of test cases. For each test case:

There is only one integer N (1 <= N <= 1018).

Output

For each test case, output the largest generalized palindromic number less than N.

Sample Input

4
12
123
1224
1122

Sample Output

11
121
1221
1121

Author: LIN, Xi                                         Source: The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

转自:http://blog.csdn.net/u011345136/article/details/39122741

题意:求小于N的回文数,这个数的回文相同的数可以缩成一个数
思路:dfs(l, r, flag):l代表左边侧长度,r代表右边的长度,flag代表是否处于边界,然后在搜索右边匹配左边的同时,枚举k,k代表右边连续相同的数都匹配左边的个数

 #include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map> #define N 55
#define M 15
#define mod 6
#define mod2 100000000
#define ll long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b) using namespace std; int T;
ll n;
char s[N];
ll ans;
int len;
char leftnum[N];
char rightnum[N]; void ini()
{
// scanf("%lld",&n);
cin>>n;
ans=-;
//memset(left,0,sizeof(left));
//memset(right,0,sizeof(right));
sprintf(s+,"%lld",n);
//sprintf(s+1,"%I64d",n);
len=strlen(s+);
for(int i=;i<=len;i++){
s[i]-='';
}
} void out()
{
//printf("%lld\n",ans);
cout<<ans<<endl;
} ll getans(int l,int r)
{
ll re=;
int i;
for(i=;i<=l;i++){
re=re*+leftnum[i];
}
for(i=r;i>=;i--){
re=re*+rightnum[i];
}
return re;
} ll dfs(int l,int r,int flag)
{
ll re=-;
int i,k,m;
if(l+r->len) return -1ll;
if(l+r-==len){
re=getans(l-,r);
if(re>=n) return -1ll;
return re;
} m= (flag==) ? s[l] : ;
for(i=m;i>=;i--)
{
leftnum[l]=i;
if( (l== || leftnum[l]!=leftnum[l-]) && !(l== && i==) && (l+r!=len) )
{
for(k=;k+l+r<=len;k++){
rightnum[k+r]=i;
re=max(re,dfs(l+,r+k,flag && (m==i) ) );
}
}
else{
re=max(re,dfs(l+,r,flag && (m==i) ));
}
if(re>){
return re;
}
}
return re;
} int main()
{
//freopen("data.in","r",stdin);
scanf("%d",&T);
for(int cnt=;cnt<=T;cnt++)
// while(T--)
//while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)
{
ini();
ans=dfs(,,);
out();
} return ;
}

ZOJ - 3816 Generalized Palindromic Number dfs的更多相关文章

  1. [ACM] ZOJ 3816 Generalized Palindromic Number (DFS,暴力枚举)

    Generalized Palindromic Number Time Limit: 2 Seconds      Memory Limit: 65536 KB A number that will ...

  2. PAT 甲级 1019 General Palindromic Number(简单题)

    1019. General Palindromic Number (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN ...

  3. General Palindromic Number (进制)

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

  4. Palindromic Number (还是大数)

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

  5. PAT1019:General Palindromic Number

    1019. General Palindromic Number (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN ...

  6. 1024 Palindromic Number int_string转换 大整数相加

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

  7. PAT A1024 Palindromic Number (25 分)——回文,大整数

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

  8. PAT A1019 General Palindromic Number (20 分)——回文,进制转换

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

  9. A1019. General Palindromic Number

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Nu ...

随机推荐

  1. OpenCV2:总结篇 core模块

    一.cv::Mat 1.作用 cv::Mat表示图像类,用来操作图像和矩阵,它包含很多属性和方法 2.构造方法 cv::Mat image;  //cv::Mat image()      无参数构造 ...

  2. servlet多文件上传(带进度条)

    需要commons-fileupload-1.3.jar和commons-io-2.4.jar的支持 页面效果:(图片文件都可以) (1)进度标识类 public class UploadStatus ...

  3. 移动产品设计之ios系统的导航

    做道题:[不定项选择题] OS中导航设计模式有几种? A.平铺导航 B.标签导航 C.树形导航 D.模态视图导航 正确答案:A B C 讲解: 导航始终是产品设计的重头戏,往往产品设计中90%的事情就 ...

  4. 常用JavaScript正则表达式整理

    在表单验证中,正则表达式书写起来特别繁琐,本文整理了15个常用的JavaScript正则表达式,其中包括用户名.密码强度.整数.数字.电子邮件地址(Email).手机号码.身份证号.URL地址. IP ...

  5. C++_STL基础案例

    C++ C++三种容器:list.vector和deque的区别:https://blog.csdn.net/gogokongyin/article/details/51178378 一.容器 小常识 ...

  6. Ubuntu美化

    Ubuntu美化 觉得ubuntu18.04的界面太丑了,所以决定美化一下. 整了好长时间特别费事.所以写个随笔记录一下. 安装gnome-tweak-tool和gnome-shell-extensi ...

  7. 【tarjan 拓扑排序 dp】bzoj1093: [ZJOI2007]最大半连通子图

    思维难度不大,关键考代码实现能力.一些细节还是很妙的. Description 一个有向图G=(V,E)称为半连通的(Semi-Connected),如果满足:?u,v∈V,满足u→v或v→u,即对于 ...

  8. 自动化运维工具Ansible

    一.简介 当下有许多的运维自动化工具( 配置管理 ),例如:Ansible.SaltStack.Puppet.Fabric 等. Ansible 一种集成 IT 系统的配置管理.应用部署.执行特定任务 ...

  9. Mining of Massive Datasets-1

    given lots of data->discover patterns and models that are: valid, useful, unexpected, understanda ...

  10. I2C驱动框架(三)

    参考:I2C子系统之platform_device初始化——smdk2440_machine_init() I2C驱动框架还应用了另一种总线-设备-驱动模型,平台设备总线platform_bus_ty ...