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

Input

Each input set will start with apositive integerN (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 shouldbe a one line output containing the summation. The value will fit in 64-bitunsigned integer.

SampleInput Outputfor Sample Input

Problemsetter: Md. Kamruzzaman

Special Thanks: Shahriar Manzoor

题意：输入n个数字，这些数字任何一种排列都是一个数，求所有整数的和

思路：每个数字固定在一位的话，那么其他位的排列数*这个数，就是这个数可能产生的数的大小，那么其实每个数都是一样的，最后再把它算到每一位上

#include 
  
   
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     #include 
    
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       typedef unsigned long long ll; using namespace std; const int maxn = 20; int num[maxn], f[maxn]; int n; void init() { f[0] = f[1] = 1; for (int i = 2; i < maxn; i++) f[i] = f[i-1] * i; } ll cal(int x) { ll tmp = 1; for (int i = 0; i < 10; i++) if (i == x) tmp *= f[num[i]-1]; else tmp *= f[num[i]]; return (ll)(f[n-1]) / tmp; } int main() { int a; init(); while (scanf("%d", &n) != EOF && n) { memset(num, 0, sizeof(num)); for (int i = 0; i < n; i++) { scanf("%d", &a); num[a]++; } ll sum = 0; for (int i = 0; i < 10; i++) if (num[i]) sum += (ll)(i) * cal(i); ll ans = 0; for (int i = 0; i < n; i++) ans = ans * 10 + sum; printf("%lld\n", ans); } return 0; }