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

 

 

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define LL unsigned long long

 4 

 5 int a[10];// 题目没有说很清楚，是0-9之间的数；

 6 int b[15];

 7 LL jie[15];

 8 int N;

 9 void init()

10 {

11     jie[0]=1;

12     for(LL i=1;i<15;i++)

13     {

14         jie[i]=i*jie[i-1];

15     }

16 }

17 LL chsort(LL x)

18 {

19     LL cnt=1;

20     for(int i=0;i<10;i++)

21     {

22         if(a[i])

23         {

24             if(i==x)

25                 cnt*=jie[a[i]-1];

26             else

27                 cnt*=jie[a[i]];

28         }

29     }

30     return cnt;

31 }

32 int main()

33 {

34    // freopen("Add.txt","r",stdin);

35 

36     LL ans,sum;

37     init();

38     while(scanf("%d",&N),N)

39     {

40         sum=0;

41         memset(a,0,sizeof(a));

42         for(int i=0;i<N;i++)

43         {

44             int tp;

45             scanf("%d",&tp);

46             a[tp]++;

47  //           sum+=b[i];

48         }

49         ans=0;

50  //       ans=jie[N-1]*sum;

51         for(LL i=0;i<10;i++)

52         {

53             if(a[i])

54             {

55                 ans+=jie[N-1]*i/chsort(i);

56             }

57         }

58         LL kk=0;

59         for(int i=1;i<=N;i++)

60         {

61             kk=kk*10+ans;

62         }

63         printf("%llu\n",kk);

64     }

65     return 0;

66 }