Add Again (sorting of repeated elements) UVA11076, adduva11076

Add Again (sorting of repeated elements) UVA11076, adduva11076
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 shocould 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

 

Train of Thought: for the nth digit, there are k Elements in total, where the nth element has ni, and the number of all elements is required: the number of all elements = (n-1 )! /(N1! * N2! * N3! * N4 !.. (Ni-1 )!.. * Nk !) Calculate the number of times each number appears in each position, and multiply by I to add up to the contribution value of the I element. Reprinted. Please indicate the source: Search for the child

 

 

1 # include <stdio. h> 2 # include <string. h> 3 # define LL unsigned long 4 5 int a [10]; // The question is not clear, it is the number between 0-9; 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 else27 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 ()); 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}