【Codeforces 1114C】Trailing Loves (or L'oeufs?)
【链接】 我是链接,点我呀:)
【题意】
【题解】
证明:https://blog.csdn.net/qq_40679299/article/details/81167283 这题的话m比较大, 做个质因数分解就ok>_< 算n!有多少个x因子的话 以5为例子 (n=25) 25 20 15 10 5 把他们都除5 5 4 3 2 1 然后再除5 1 所以总共有6个 转换成代码就是 while(n>0){ ans+=n/5; n = n/5; }【代码】
import java.io.*;
import java.util.*;
public class Main {
static int N = (int)1e6;
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static class Task{
public void solve(InputReader in,PrintWriter out) {
long n,m;
n = in.nextLong();m = in.nextLong();
ArrayList a = new ArrayList<>();
ArrayList d = new ArrayList<>();
for (long i = 2;i*i<=m;i++) {
if (m%i==0) {
a.add(i);
int cnt1 = 0;
while (m%i==0) {
m/=i;
cnt1++;
}
d.add(cnt1);
}
}
if (m>1) {
a.add(m);
d.add(1);
}
long ans = (long)1e18 + 100;
for (int i = 0;i <(int)a.size();i++) {
long ai = (long) a.get(i);
int pi = (int) d.get(i);
long nn = n;
long res = 0;
while (nn>0) {
res+=nn/ai;
nn/=ai;
}
ans = Math.min(ans, res/pi);
}
out.print(ans);
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}