HDU 3304 calculates the number of integers divisible by prime p in row N of the Yang Hui triangle.

Equivalent to C (n, m) % P = 0. Evaluate the number of M.

N! /(M! * (N-m )!) % P = 0

N! The number of P factors equals to M! And (n-m )! Factors and

That is, F (n) = f (m) + f (n-m) // F (n) indicates n! Number of factors contained

F (n) = N/P + N/(P ^ 2) + ....

That is:

N/P + N/P ^ 2 +... = m/P + M/P ^ 2 + .. + (n-M)/P ^ 2 +... (1)

Because N/P ^ I> = m/P ^ I + (n-M)/P ^ I,

Therefore, it is equivalent to N/P ^ I = m/P ^ I + (n-M)/P ^ I (2) for any I)

N = N/P ^ I * P ^ I + N % P ^ I (note that all division here is Division)

(2) Both sides are multiplied by P ^ I

N-N % P ^ I = m-m % P ^ I + N-M + (n-m) % P ^ I

N % P ^ I-m % P ^ I = (n-m) % P ^ I> = 0

It is equivalent to M % (P ^ I) <= n % (P ^ I) (3) for any positive integer I. (// This step from 2 to 3 is very important and is a key point of this question)

Write N and M as numbers N0, N1, N2, N3, N4,... M0, M1, M2, M3, M4 ,....

 

According to (3) it can be obtained by mathematical induction. For any bit, Ni is greater than or equal to Mi.

 

When I = 0; apparently true

Assume that I = K is true. When I = k + 1

Because M % (P ^ K) <= n % (P ^ K)

N (k + 1) nk .. n1 = N (k + 1) (NK % (P ^ K) (indicating placement by P in hexadecimal format)

Therefore, ensure that N % (P ^ (k + 1)> = m % (P ^ (k + 1 ))

The maximum value must be n (k + 1)> = m (k + 1)

 

That is, (3) the necessary and sufficient conditions are M0 <= N0, M1 <= N1 ....

 

So the value of M is (N0 + 1) * (N1 + 1) * (n2 + 1 )......

 

•:

// HDU 3304

# include
using namespace STD;
int main ()
{< br> int N, P, ANS, Cs;
cs = 0;
while (CIN> P> N)
{< br> CS ++;
If (n = 0) & (P = 0) break;
ans = 1;
while (n! = 0)
{< br> ans = ans * (N % P + 1) % 10000;
N = N/P;
}< br> cout <"case" If (ANS <1000) cout <'0 ';
If (ANS <100) cout <'0';
If (ANS <10) cout <'0 ';
the cout }< BR >}