Question: Seek the number of suffixes 0 in the k system.

The old man pinched his finger, even pinched the std algorithm?

• Solution
  Reasonably, at first sight, this problem will not be done.
  But the old man thought that if m=10, he would still do it.
  If the number is less than or equal to n, how many multiples of 5 will be enough. 2 is absolutely more than 5.
  So I had an idea.
  First, prime factor decomposition is indispensable. m=ki=1scntii
  Then, for each prime factor si, calculate how many multiples [1,n] can find (but when you encounter 23 in a similar calculation of 2, these are counted three times)
  How to calculate? Let's face it? Should I be the only zz to think that way at the beginning?
  Later, it was found that adding 1 to each calculation was enough, because the previous calculation had already been done.
  After calculating, it is found that si finds xi in [1,n], but the occurrence of at least one m si requires cnti times. So the final answer is minki=1{xicnti}

#include <bits/stdc++.h>
#define LL long long
using namespace std ;
LL n, m, tot ;
LL s[1010], cnt[1010] ;
int main() {
#ifndef ONLINE_JUDGE
    freopen ( "a.in", "r", stdin ) ;
    freopen ( "a.out", "w", stdout ) ;
#endif
    LL i, j, x, ans, gn ;
    while ( scanf ( "%lld%lld", &n, &m ) != EOF ) {
        memset ( s, 0, sizeof s ) ;
        memset ( cnt, tot = 0, sizeof cnt ) ;
        for ( x = m, i = 2 ; i*i <= x ; i ++ ) {
            if (x%i == 0) s[++tot] = i ;
            while (x%i == 0) cnt[tot] ++, x /= i ;
        }
        if (x != 1) s[++tot] = x, cnt[tot] = 1 ;    
        /*printf ( "%lld = ", m ) ;
        for ( i = 1 ; i < tot ; i ++ ) printf ( "%lld^%lld + ", s[i], cnt[i] ) ;
        printf ( "%lld^%lld\n", s[tot], cnt[tot] ) ;*/
        ans = 1LL<<60 ;
        for ( i = 1 ; i <= tot ; i ++ ) {
            gn = 1 ;    
            x = 0 ; 
            for ( j = 1 ; gn <= n ; j ++ ) {
                gn *= s[i] ;
                x += n/gn ;
            }
            //printf ( "time %lld = %lld\n", s[i], x ) ;    
            ans = min(ans, x/cnt[i]) ;  
        }
        printf ( "%lld\n", ans ) ;  
    }
    return 0 ;
}

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