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Given are two integers (x and n)  you have to find an array such that it contains the frequency of index numbers occurring in ( x^1 x^2 ....x^(n-1) x^(n) ) . 

Input:
The first line of input contains an integer "test" denoting the number of test cases . Then "test" test cases follow . The first line of each test case contains two space separated integers x and n, where x is an integer (1 <= x <= 15) and n is the number of times to which x is to be raised  ,  in increasing manner .  x^1  x^2  x^3 ....... x^(n-1)  x^(n) .


Output: 
Corresponding to each test case, in a new line ,  print an array "a[]" which containes the frequency of each digit occuring in input (frequency of '0' at index "0" , frequency of '1' at index "0" ........ frequency of '9' at index "9" )  .

 

Constraints  :
1<=T<=100
1 < = x < = 15
1 < = n < = 10


Example: 
Input :
5  
15  3  
2 4    
3 2
1  5
5 2   
Output : 
0 1 2 2 0 3 0 1 0 0
0 1 1 0 1 0 1 0 1 0
0 0 0 1 0 0 0 0 0 1

0 5 0 0 0 0 0 0 0 0
0 0 1 0 0 2 0 0 0 0  

Explanation :
Example 1: 

First testcase:  15 3 
15^1 15^2 15^3 ==> 15, 225, 3375 = s

Output :
An array which displays the frequency of its index numbers occuring in s = 15 225 3375
Frequency of ' 0 ' = 0
Frequency of ' 1 ' = 1 (only once in s , in 15)
Frequency of ' 2 ' = 2 (twice in s , in 225)
Frequency of ' 3 ' = 2 (twice in s , in 3375 )
Frequency of ' 4 ' = 0 ( no where in s)
Frequency of ' 5 ' = 3 ( thrice in s , in '15' , in '225' ,  in '3375' )
Frequency of ' 6 ' = 0
Frequency of ' 7 ' = 1  (only once in s , in '3375' )
Frequency of ' 8 ' = 0
Frequency of ' 9 ' = 0

Output array : a[] = 
a[0] = 0
a[1] = 1
a[2] = 2
a[3] = 2
a[4] = 0
a[5] = 3
a[6] = 0
a[7] = 1
a[8] = 0
a[9] = 0

Resultant array:
0 1 2 2 0 3 0 1 0 0

** For More Input/Output Examples Use 'Expected Output' option **

Contributor: Diksha Singhal

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