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Input: Standard Input

Output: Standard Output

Summation of sequence of integers is always a common problem in Computer Science. Rather than computing blindly, some intelligent techniques make the task simpler. Here you have to find the summation of a sequence of integers. The sequence is an interesting one and it is the all possible permutations of a given set of digits. For example, if the digits are <1 2 3>, then six possible permutations are <123>, <132>, <213>, <231>, <312>, <321> and the sum of them is 1332.

Input

Each input set will start with a positive integer N (1≤N≤12). The next line will contain N decimal digits. Input will be terminated by N=0. There will be at most 20000 test set.

Output

For each test set, there should be a one line output containing the summation. The value will fit in 64-bit unsigned integer.

Sample Input         Output for Sample Input

題意：給你n個數字（0~9，1<=n<=12）,問這些數字構成的所有不重複排列的和。

分析：舉個例子

含重複數字時能構成的所有不重複排列的個數為：(n!)/((n1!)*(n2!)*...*(nn!))，其中ni指數字i出現的次數。

又因為每個數字在每一位出現的概率時等可能的。

比如1 1 2，所能構成的所有情況為

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

而1、2、3出現在個、十、百位的次數時一樣的，即6/3;

則每個數字在每一位出現的次數為 [(n!)/((n1!)*(n2!)*...*(nn!))]/n;（含重複數字時同樣適用）

簡化加法，即每個數字在每一位均出現1次時這個數字的和為 x*1...1 (n個1)

則n個數字在每一位出現times次，即為所求答案。ans = (a1+a2+...+an)*(1...1)*[(n!)/((n1!)*(n2!)*...*(nn!))]/n;

切忌：[(n!)/((n1!)*(n2!)*...*(nn!))]/n*(a1+a2+...+an)*(1...1)這樣表達時錯誤的，當n個數字相同時，[(n!)/((n1!)*(n2!)*...*(nn!))] = 1, 1/n會得到0，所以應先乘再除；

【程式碼】：

 #include<iostream>

 #include<cstdio>

 #include<cstdlib>

 #include<cstring>

 using namespace std;

 typedef unsigned long long ull;

 const int maxn = ;

 int x, a[maxn], num[maxn];

 ull C[maxn];

 const ull basic[] =

 {

     , , , , , , , ,

     , , ,

 };

 void init()

 {

     C[] = C[] = ;

     for(int i = ; i <= ; i++)

     {

         C[i] = C[i-]*i;

     }

 }

 int main()

 {

     init();

     int n;

     while(scanf("%d", &n) && n)

     {

         memset(num, , sizeof(num));

         ull ans = ;

         for(int i = ; i < n; i++)

         {

             scanf("%d", &x);

             ans += x;

             num[x]++;

         }

         ull times = C[n];

         for(int i = ; i < ; i++)

         {

             times /= C[num[i]];

         }

         ans = ans*times*basic[n-]/n;

         cout << ans << endl;

     }

     return ;

 }