Lweb and String
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 368 Accepted Submission(s): 243
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)
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.
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
2
aabcc
acdeaa
Sample Output
Case #1: 3
Case #2: 4
Author
UESTC
Source
解析:说了那么多,其实很简单,就是求有多少种字母。
#include <cstdio> #include <cstring> const int MAXN = 1e5+5; char s[MAXN]; bool vis[30]; int main() { int t, cn = 0; scanf("%d", &t); while(t--){ scanf("%s", s); memset(vis, 0, sizeof(vis)); int res = 0; for(int i = 0; s[i] != ' '; ++i){ if(!vis[s[i]-'a']){ vis[s[i]-'a'] = true; ++res; } } printf("Case #%d: %d ", ++cn, res); } return 0; }