Nyist 56 factorial decomposition (1)

Factorial Factorization (I) time limit: 3000 MS | memory limit: 65535 kb difficulty: 2

Description

Given two numbers m, n, where M is a prime number.

Returns the factorial of N (0 <= n <= 10000) to calculate the number of M.

 

Input

The first row is an INTEGER (0 <S <= 100), indicating the number of test data groups.
The next s row has two integers n and M.

Output

Number of output M.

Sample Input

2100 516 2

Sample output

2415

 

 

 

# Include <stdio. h>
Int main ()
{
Int S1, n, m, S, J, I, II;
Scanf ("% d", & S1 );
While (S1 --)
{S = 0;
Scanf ("% d", & N, & M );
For (I = 1; I <= N; I ++)
{II = I;
If (II % m = 0)
{
S ++; II = II/m;

While (II % m = 0)
{
S ++; II = II/m;
}
}
}
Printf ("% d \ n", S );
}
}

 

 

# Include <iostream>
Using namespace STD;
Int main ()
{
Int m, n, k, S, sum;
Cin> S;
While (s --)
{
Sum = 0;
Cin> N> m; // M is a prime number.
While (N)
{
K = N/m;
Sum + = K;
N = K;
}
Cout <sum <Endl;
}
}

// Factorial decomposition. I thought it would be n! The number of M in all factorial values.
// This is not the original meaning.
// The description of the question is incorrect.
// 100/5 = 20 5
// 20/5 = 4 5
// 4/5 = 0 5
// End
// 20 + 4 + 0 = 24

 

**************************************** ***********

8 *************************************** **************************************** ******

 

# Include <stdio. h>
Int main ()
{
Int count, n, m, I, ii, T;

Scanf ("% d", & T );
While (t --)
{
Count = 0;
Scanf ("% d", & N, & M );
For (I = 1; I <= N; I ++)
{II = I;
If (II % m = 0) // check one by one
{
Count ++; // + 1
II = II/m; // whether the new II contains a certain power of m

While (II % m = 0) // a power of m
{Count ++; II = II/m ;}//
}


}
Printf ("% d", count );
}

}