hdu5842-Lweb and String
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Lweb and String
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1295 Accepted Submission(s): 655
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 injectionf: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 fromS .
LIS: Longest Increasing Subsequence. (https://en.wikipedia.org/wiki/Longest_increasing_subsequence)
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.
Every injection
For example, a String “aabc”,
Now help Lweb, find the longest LIS which you can obtain from
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) .
ThenT lines follow, the i-th line contains a string S only containing the lowercase letters, the length of S will not exceed 105 .
Then
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
2aabccacdeaa
Sample Output
Case #1: 3Case #2: 4题目大意:给出n个字符串,每个字符串皆由小写字母组成,输出每个字符串含有的不同字符的个数#include<iostream>#include<algorithm>#include<cstring>#include<string>using namespace std;int main(){ int n; cin>>n; int kase=0; while(n--) { string s; cin>>s; int num=0; int len=s.length(); int a[30]; memset(a,0,sizeof(a)); for(int i=0;i<len;i++) { if(a[s[i]-'a']==1)continue; else { a[s[i]-'a']=1; num++; } } cout<<"Case #"<<++kase<<": "<<num<<endl; } return 0;}
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