Va 10139 factovisors (number theory)

Factovisorsthe factorial function, n! Is defined thus for n a non-negative integer:

   0! = 1   n! = n * (n-1)!   (n > 0)

We say that a divides B if there exists an integer k such that

   k*a = b

The input to your program consists of several lines, each containing two non-negative integers, N and M, both less than 2 ^ 31. for each input line, output a line stating whether or not m divides n !, In the format shown below.

Sample Input

6 96 2720 1000020 1000001000 1009

Output for sample input

9 divides 6!27 does not divide 6!10000 divides 20!100000 does not divide 20!1009 does not divide 1000!

Question: Give N and M, and ask if m can divide the factorial of N. Analysis: prime factor decomposition can be performed on m to obtain the number of each prime factor, and n! To compare the number of this factor. If it is greater than n! The number of this factor, you cannot divide.

# Include <stdio. h> # include <string. h> # include <math. h> const int maxn = 100005; int vis [maxn], prime [10000], num; void get_prime () // returns the prime number {num = 0; memset (VIS, 0, sizeof (VIS); For (INT I = 2; I <maxn; I ++) {If (! Vis [I]) {Prime [num ++] = I; for (Int J = I + I; j <maxn; j + = I) vis [J] = 1 ;}}} int CAL (int w, int p) // calculate the number of P in the factorial of W {int ans = 0; while (W) {w/= P; ans + = W;} return ans;} bool judge (int n, int m) {int K = (INT) SQRT (m + 0.5 ); for (INT I = 0; I <num & prime [I] <= K; I ++) {If (M % prime [I] = 0) {int CNT = 0; while (M % prime [I] = 0) {CNT ++; M/= prime [I];} If (CAL (n, prime [I]) <CNT) return false ;} // If m at this time! = 1, then M must be a prime number. If n> = m, m must be able to divide n! If (M> 1 & n <m) return false; return true;} int main () {int n, m; get_prime (); While (~ Scanf ("% d", & N, & M) {If (Judge (n, m) printf ("% d divides % d! \ N ", m, n); else printf (" % d does not divide % d! \ N ", m, n);} return 0 ;}