strcmp() is a library function in C/C++ which compares two strings. It takes two strings as input parameter and decides which one is lexicographically larger or smaller: If the first string is greater then it returns a positive value, if the second string
is greater it returns a negative value and if two strings are equal it returns a zero. The code that is used to compare two strings in C/C++ library is shown below:

The number of comparisons required to compare two strings in strcmp() function is never returned by the function. But for this problem you will have to do just that at a larger scale. strcmp() function continues to compare characters in the same position
of the two strings until two different characters are found or both strings come to an end. Of course it assumes that last character of a string is a null (‘\0’) character. For example the table below shows what happens when “than” and “that”; “therE” and
“the” are compared using strcmp() function. To understand how 7 comparisons are needed in both cases please consult the code block given above.

Input

The input file contains maximum 10 sets of inputs. The description of each set is given below:

Each set starts with an integer N (0<N<4001) which denotes the total number of strings. Each of the next N lines contains one string. Strings contain only alphanumerals (‘0’… ‘9’, ‘A’… ‘Z’, ‘a’… ‘z’) have a maximum length of 1000, and a minimum length of
1.  

Input is terminated by a line containing a single zero. Input file size is around 23 MB.

Output

For each set of input produce one line of output. This line contains the serial of output followed by an integer T. This T denotes the total number of comparisons that are required in the strcmp() function if all the strings are compared with one another
exactly once. So for N strings the function strcmp() will be called exactly  times. You have to calculate total number of comparisons inside the strcmp() function in
those  calls. You can assume that the value of T will fit safely in a 64-bit signed integer. Please note that the most straightforward solution (Worst Case Complexity
O(N² *1000)) will time out for this problem.

Sample Input                              Output for Sample Input

详见大白书：p210

每个节点记录两个域：

1.从该节点下去的串数。

2.该节点的结束标识。

做法：

1.一开始就把0节点 jud 加上。相当于是把所有不与现在插入串  匹配的串加上那个不匹配的判断 即 s[ i ] != t[ i ]

2. 然后匹配一个就把 答案 加上2次jud

3. 如果完全匹配的串 要加上最后的 结尾数。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstdlib>

using namespace std;

#define MAXN 4010*1000
#define MAX 65

int sz,t[MAXN][MAX];
int jud[MAXN];
int fin[MAXN];

void clear(){
    sz=1;
    memset(t[0],-1,sizeof(t[0]));
    jud[0]=0;
    fin[0]=0;
}

int idx(char c){
    if(c<='9'&&c>='0')
        return c-'0';
    if(c<='Z'&&c>='A')
        return c-'A'+11;
    return c-'a'+37;
}

long long ret;

void insert(char *s){
    int u=0;
    int n=strlen(s);
    ret+=jud[0];
    jud[0]++;
    for(int i=0;i<n;i++){
        int c=idx(s[i]);
       // cout<<t[u][c]<<" "<<u<<" "<<c<<endl;
        if(t[u][c]==-1){
            memset(t[sz],-1,sizeof(t[sz]));
            jud[sz]=0;
            fin[sz]=0;
            t[u][c]=sz++;
        }
        u=t[u][c];
        ret+=jud[u];
        ret+=jud[u];
        jud[u]++;
    }
    ret+=fin[u];
    fin[u]++;
}

int n;
char str[2000];

int main(){
    int cs=1;
    while(~scanf("%d",&n)&&n){
        ret=0;
        clear();
        for(int i=0;i<n;i++){
            scanf("%s",str);
            insert(str);
        }
        printf("Case %d: %lld\n",cs++,ret);
    }
    return 0;
}

/**
3
that
than
thaa
4
a1At
jtt
a1At
a12b
a12b
**/