UVA10780 exponentiation and factorial

Test instructions
Enter two integers n,m (1<m<5000,0<n<10000) to find the smallest k so that m^k is the n! factor.


Ideas:
Relatively easy to think, at the beginning of the hands of the WA several times, we directly find M and n! prime factor and number on the line, if S1[a] represents the number of prime factor a n!, S2 is M, then Ans=min (Ans, s1[a]/s2[a]); This should not be explained, well understood.




#include <stdio.h>
#include <string.h>


int pri[11000], pt;
int mark[11000];
int s1[11000], s2[11000];


void Dbpri ()
{
memset (mark, 0, sizeof (Mark));
MARK[1] = 1;
PT = 0;
for (int i = 2; I <= 10000; i + +)
{
if (!mark[i])
{
PRI[++PT] = i;
for (int j = i + i; J <= 10000; j + = i)
MARK[J] = 1;
}
}
}


int main ()
{
Dbpri ();
int T, CAS = 1, I, J, N, M;
scanf ("%d", &t);
while (t--)
{
scanf ("%d%d", &m, &n);
memset (S1, 0, sizeof (S1));
for (i = 1; I <= n; i + +)
{
int now = i;
for (j = 1; PRI[J] <= now && J <= pt; j + +)
{
while (now% pri[j] = = 0)
{
Now/= Pri[j];
S1[PRI[J]] + +;
}
}
}




memset (s2, 0, sizeof (S2));
int mm = m;
for (i = 1; Pri[i] <= mm && i <= pt; i + +)
if (mm% pri[i] = = 0)
{
while (mm% pri[i] = = 0)
{
S2[pri[i]] + +;
MM/= pri[i];
}
}
int Ans = 100000;
for (i = 1; Pri[i] <= m && i <= pt; i + +)
if (m% pri[i] = = 0)
{
if (Ans > S1[pri[i]]/s2[pri[i])
Ans = S1[pri[i]]/s2[pri[i]];
}


printf ("Case%d:\n", CAS + +);
if (Ans = = 0) printf ("Impossible to divide\n");
else printf ("%d\n", Ans);
}
return 0;
}