Add Again(重复元素排序) UVA11076

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
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 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

---

Problemsetter: Md. Kamruzzaman

Special Thanks: Shahriar Manzoor

思路：对于第x位，一共有k个元素,其中第i个元素有ni个,求全排列个数：全排列个数=（n-1）!/(n1!*n2!*n3!*n4!..(ni-1)!..*nk!)算出每个数出现在每个位置的次数，然后乘以i加起来就是i元素的贡献值了

转载请注明出处：寻找&星空の孩子

 #include<stdio.h>
 #include<string.h>
 #define LL unsigned long long
 
 int a[];// 题目没有说很清楚，是0-9之间的数；
 int b[];
 LL jie[];
 int N;
 void init()
 {
     jie[]=;
     for(LL i=;i<;i++)
     {
         jie[i]=i*jie[i-];
     }
 }
 LL chsort(LL x)
 {
     LL cnt=;
     for(int i=;i<;i++)
     {
         if(a[i])
         {
             if(i==x)
                 cnt*=jie[a[i]-];
             else
                 cnt*=jie[a[i]];
         }
     }
     return cnt;
 }
 int main()
 {
    // freopen("Add.txt","r",stdin);
 
     LL ans,sum;
     init();
     while(scanf("%d",&N),N)
     {
         sum=;
         memset(a,,sizeof(a));
         for(int i=;i<N;i++)
         {
             int tp;
             scanf("%d",&tp);
             a[tp]++;
  //           sum+=b[i];
         }
         ans=;
  //       ans=jie[N-1]*sum;
         for(LL i=;i<;i++)
         {
             if(a[i])
             {
                 ans+=jie[N-]*i/chsort(i);
             }
         }
         LL kk=;
         for(int i=;i<=N;i++)
         {
             kk=kk*+ans;
         }
         printf("%llu\n",kk);
     }
     return ;
 }