Problem Description

Lweb has a string S.

Oneday, he decided to transform this string to a new sequence.

You need help him determine this transformation to get a sequence which has the longest LIS(Strictly Increasing).

You need transform every letter in this string to a new number.

A is the set of letters of S, B is the set of natural numbers.

Every injection f:A→B can be treat as an legal transformation.

For example, a String “aabc”, A={a,b,c}, and you can transform it to “1 1 2 3”, and the LIS of the new sequence is 3.

Now help Lweb, find the longest LIS which you can obtain from S.

LIS: Longest Increasing Subsequence. (https://en.wikipedia.org/wiki/Longest_increasing_subsequence)

Input
The first line of the input contains the only integer T,(1≤T≤20).

Then T lines follow, the i-th line contains a string S only containing the lowercase letters, the length of S will not exceed 105.

 
Output
For each test case, output a single line "Case #x: y", where x is the case number, starting from 1. And y is the answer.
 
Sample Input

aabcc
acdeaa Sample Output
Case #:
Case #:

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5842

***************************************************

题意:我只能说题意神马都是浮云,纯属干扰,直接看代码吧

 #include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
#define N 1100 char s[];
int v[]; int main()
{
int i,T,k=,len; scanf("%d", &T); while(T--)
{
memset(v, , sizeof(v));
int ans=;
scanf("%s",s);
len=strlen(s); for(i=;i<len;i++)
v[s[i]-'a']=; for(i=;i<;i++)
if(v[i])
ans++; printf("Case #%d: %d\n", k++,ans);
}
return ;
}

2016中国大学生程序设计竞赛 - 网络选拔赛 1011 Lweb and String的更多相关文章

  1. 2016中国大学生程序设计竞赛 - 网络选拔赛 C. Magic boy Bi Luo with his excited tree

    Magic boy Bi Luo with his excited tree Problem Description Bi Luo is a magic boy, he also has a migi ...

  2. 2016中国大学生程序设计竞赛 - 网络选拔赛 J. Alice and Bob

    Alice and Bob Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  3. 2016中国大学生程序设计竞赛 网络选拔赛 I This world need more Zhu

    This world need more Zhu Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Jav ...

  4. 2016中国大学生程序设计竞赛 - 网络选拔赛 1004 Danganronpa

    Problem Description Chisa Yukizome works as a teacher in the school. She prepares many gifts, which ...

  5. 2016中国大学生程序设计竞赛 - 网络选拔赛 1001 A water problem (大数取余)

    Problem Descripton Two planets named Haha and Xixi in the universe and they were created with the un ...

  6. 2018中国大学生程序设计竞赛 - 网络选拔赛 1001 - Buy and Resell 【优先队列维护最小堆+贪心】

    题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6438 Buy and Resell Time Limit: 2000/1000 MS (Java/O ...

  7. 2018中国大学生程序设计竞赛 - 网络选拔赛 1010 YJJ's Salesman 【离散化+树状数组维护区间最大值】

    题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6447 YJJ's Salesman Time Limit: 4000/2000 MS (Java/O ...

  8. 2018中国大学生程序设计竞赛 - 网络选拔赛 1009 - Tree and Permutation 【dfs+树上两点距离和】

    Tree and Permutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Oth ...

  9. HDU 6154 - CaoHaha's staff | 2017 中国大学生程序设计竞赛 - 网络选拔赛

    /* HDU 6154 - CaoHaha's staff [ 构造,贪心 ] | 2017 中国大学生程序设计竞赛 - 网络选拔赛 题意: 整点图,每条线只能连每个方格的边或者对角线 问面积大于n的 ...

随机推荐

  1. sql分页带参数,带排序等,动态实现的方法

    USE [YQOBS] GO /****** Object: StoredProcedure [dbo].[PageList] Script Date: 11/06/2014 11:39:35 *** ...

  2. runtime官方文档

    OC是一种面向对象的动态语言,作为初学者可能大多数人对面向对象这个概念理解的比较深,而对OC是动态语言这一特性了解的比较少.那么什么是动态语言?动态语言就是在运行时来执行静态语言的编译链接的工作.这就 ...

  3. c/c++ 浮点型处理

    #include <stdio.h> #include <iostream> #include <string> #include <string.h> ...

  4. 测试sql性能方法

    SET STATISTICS io ON         SET STATISTICS time ON         go          ---你要测试的sql语句          selec ...

  5. oracle导入导出数据库

    oracle导出dmp文件: 开始->运行->输入cmd->输入 exp user/password@IP地址:1521/数据库实例 file=文件所在目录 (如:exp user/ ...

  6. Ip 讲解

    IP地址分类以及C类IP地址的子网划分 国际规定:把所有的IP地址划分为 A,B,C,D,E A类地址:范围从0-127,0是保留的并且表示所有IP地址,而127也是保留的地址,并且是用于测试环回用的 ...

  7. HDU 5798 Stabilization

    方法太厉害了....看了官方题解的做法....然后...想了很久很久才知道他想表达什么.... #pragma comment(linker, "/STACK:1024000000,1024 ...

  8. BaLaBaLa

    #include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<a ...

  9. 眼睛跟踪 java

    https://github.com/hosek/eyeTrackSample Simple sample, for eye tracking with OpenCV

  10. 运维命令rsync

    如果你是一位运维工程师,你很可能会面对几十台.几百台甚至上千台服务器,除了批量操作外,环境同步.数据同步也是必不可少的技能. 说到“同步”,不得不提的利器就是rsync,今天就来说说我从这个工具中看到 ...