UVa11076

11076 Add Again
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
31
2 3
31
1 2
0
Sample Output
1332
444

题意：
       输入n个数字，这些数字的任意排列总是一个数。你的任务是求出所有这些数的和。

分析：

       其实，输入的每个数字在每一位出现的概率都是相同的，于是，我们可以先计算出所有输入数字的平均值A。而可重复的排列数就是P=n!/(n1!*n2!*…*nk!)最终的答案就是11…1(n个1)*A*P。

#include <cstdio>
#include <cstring>
#define ll long long
int N;
ll factorial[13];
int num[10];
const ll one[13] = {
    0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111,
    111111111, 1111111111, 11111111111, 111111111111
};
void cal_factorial(){
    factorial[0] = 1,factorial[1] = 1;
    for(int i = 2 ; i <= 12 ; i++) factorial[i] = i * factorial[i - 1];
}
int main(){
    cal_factorial();
    int sum = 0;
    while(scanf("%d",&N) == 1 && N){
        sum = 0;
        memset(num,0,sizeof num);
        for(int i = 0 ; i < N ; i++){
            int tmp; scanf("%d",&tmp);
            num[tmp]++,sum += tmp;
        }
        ll ans = factorial[N - 1] * sum;
        for(int i = 0 ; i < 10 ; i++) ans /= factorial[num[i]]; // 平均数需要除以N，将分子上N
        printf("%lld\n",ans * one[N]);
    }
    return 0;
}