Lweb and String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 261 Accepted Submission(s): 174

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

2
aabcc
acdeaa

Sample Output

Case #1: 3
Case #2: 4

Author

UESTC

Source

题意：给你一个字符串，你把它转换成数字，然后求最大递增子序列。

大坑，一直在求最大递增子序列，一直WA，最后发现连最大递增子序列也不用求了，算出字符串中26个小写字母有几个就行了。

代码：

#include<stdio.h>
#include<string.h>
#define N 10000006

char s[N];
int a[N];

int main(void)
{
int t, ans, i, j;

scanf("%d", &t);

j = 0;
while(t--)
{
j++;
ans = 0;
memset(a, 0, sizeof(a));
scanf("%s", s);

int n = strlen(s);

for(i = 0; i < n; i++)
a[s[i] - 'a' + 1]++;

for(i = 1; i <= 26; i++)
{
if(a[i])//判断26个英文字母是否出现过，出现过长度就加1.
ans++;

}

printf("Case #%d: %d\n", j, ans);
}

}

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