Description
Khaled was sitting in the garden under an apple tree, suddenly! , well... you should guess what happened, an apple fell on his head! , so he came up with a new Cryptography method!! The method deals only with numbers, so... If you want to encode a number, you must represent each of its digits with a set of strings, then the size of the set is the digit itself, No set should contain the same string more than once. For example: the number 42, can be represented with the following two sets: 1) "dog" "load" "under" "nice". 2) "stack" "dog". The first set contain four strings so it represent the digit 4. The second set contain two strings so it represent the digit 2. Given N strings, what is the smallest number you can get from dividing these strings into non-empty sets, and then decode the result by Khaled's Cryptography method? , You must use all the given strings, and no set should contain the same string more than once.
Input
The input consists of several test cases, each test case starts with 0 < N ≤ 10000, the number of the given strings, then follows N space-separated string, each string will contain only lower-case English letters, and the length of each string will not exceeded 100. You can assume that there are no more than nine distinct strings among the given strings. A line containing the number 0 defines the end of the input you should not process this line.
Output
For each test case print a single line in the following format: "Case c: x" where c is the test case number starting from 1 and x is the solution to the described problem above.
Sample Input
3 one two two
7 num go book go hand num num
25 aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa aa
0
Case 1: 12
Case 2: 124
Case 3: 1111111111111111111111111
Hint
In the first sample, we divided the given strings into two sets, the first set contains two word: "one" and "two" so it represents the digit 2, the second set contains only one word: "two" so it represent the digit 1.
/*/
统计输入相同字符串的个数，从小到大把字符串分组，每组内不能有相同字符串。 输出每组的字符串个数。 AC代码：
/*/
#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"cstdio"
#include"string"
#include"vector"
#include"queue"
#include"cmath"
using namespace std;
typedef long long LL ;
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x)) string s[10005]; int main() {
int n,time=1;
int num[10005],ans[10005];
while(~scanf("%d",&n)) {
if(!n)break;
memset(ans,0);
for(int i=0; i<n; i++) {
cin>>s[i];
num[i]=1;
}
sort(s,s+n);
int j=0;
for(int i=1; i<=n; i++) {
if(s[i]==s[i-1])num[j]++;
else j++;
}
int maxx=0;
for(int i=0;i<j;i++){
maxx=max(num[i],maxx);
for(int k=0;k<num[i];k++){
ans[k]++;
}
}
printf("Case %d: ",time++);
for(int i=maxx-1;i>=0;i--){
printf("%d",ans[i]);
}
puts("");
}
return 0;
}