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; }
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