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Mathematically beautiful numbers | Hackerearth practice problem solution

Rajnish Tripathi 22:47

   Problem:-

While playing a mental math game, you realize that the number k is mathematically beautiful.

You then realize that the number x can be mathematically beautiful if it is represented as a sum of a sequence where each element is a power of k and all the numbers in the sequence are different.

Task

Your task is to determine whether the number is mathematically beautiful.

Input format

  • The first line contains T denoting the number of test cases.
  • The next T lines contain x  and  k  denoting the numbers.

Output Format

For each test case, output  "YES"  if x is "mathematically beautiful" and  "NO"  otherwise.

Constraints

T<=1000

(1<=x<=1018)

(2<=k<=9)

 

Sample Input
2
91 3
17 5
Sample Output
YES
NO
Time Limit: 2
Memory Limit: 256
Source Limit:
Explanation

91=30+32+34

.....................................................

Code:-

  Here I am going to give you two solution first one is on the basis of C language and second one is on the basis of c++ language which you can submit in c++14 and c++17 also

Solution 1 ( C language):-

#include <stdio.h>
int main()
{
    int T, k, flag, i, rem;
    long long int x;
    scanf("%d", &T);
    for(i=0; i<T; i++)
    {
        scanf("%lld %d",&x , &k );
        flag = 0;
        while(x)
        {
            rem = x % k ;
            if(rem != 0 && rem != 1)
            {
                flag = 1;
                break;
            }
            x = x/k ;
        }
        if(flag == 1)
            printf("NO\n");
        else
            printf("YES\n");
    }
}

Solution 2 ( C++ language):-

 This solution is based on the c++ language and you can submit ib c++14 and c++17 also.
#include <iostream>
using namespace std;
int main()
{
int t,k;
long long int x;
cin>>t;
while(t--)
{
cin>>x>>k;
while(x!=0)
{
if(x%k!=0)
{
x--;
if(x%k!=0)
{
cout<<"NO"<<endl;
break;
}
else
x=x/k;
}
else
x=x/k;
if(x==0)
cout<<"YES"<<endl;
}
}
return 0;
}

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