【数论-数位统计】UVa 11076 - Add Again

Add Again
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

3 1 2 3 3 1 1 2 0

1332 444

题意：给你n个数字（0~9，1<=n<=12）,问这些数字构成的所有不重复排列的和。

分析：举个例子

含重复数字时能构成的所有不重复排列的个数为：(n!)/((n1!)*(n2!)*...*(nn!))，其中ni指数字i出现的次数。

又因为每个数字在每一位出现的概率时等可能的。

比如1 1 2，所能构成的所有情况为

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

而1、2、3出现在个、十、百位的次数时一样的，即6/3;

则每个数字在每一位出现的次数为 [(n!)/((n1!)*(n2!)*...*(nn!))]/n;（含重复数字时同样适用）

简化加法，即每个数字在每一位均出现1次时这个数字的和为 x*1...1 (n个1)

则n个数字在每一位出现times次，即为所求答案。ans = (a1+a2+...+an)*(1...1)*[(n!)/((n1!)*(n2!)*...*(nn!))]/n;

切忌：[(n!)/((n1!)*(n2!)*...*(nn!))]/n*(a1+a2+...+an)*(1...1)这样表达时错误的，当n个数字相同时，[(n!)/((n1!)*(n2!)*...*(nn!))] = 1, 1/n会得到0，所以应先乘再除；

【代码】：

 #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 ;
 }