n! decomposition of the element factor

n! = p1^t1*p2^t2* ... .. Pi^ti ... .. *PK^TK (where P1 , P2 ... PK is the prime number, 1<n<= 10^6 )

Obviously, we first have to sift out all primes that are less than or equal to N.

And then we're going to divide N into two parts,

1) is a multiple of a factor, 2) is not a multiple of this factor

Example: F (n,2)

n!= (2*4*6*....N) * (1*3*5*.....* (n-1))

=2^ (N/2) * (1*2*3*4*......N/2) * (1*3*5*.  (n-1)) Got a

......

So we get the number of times a recursive formula has a factor f (n,p) = f (n/p,p) + n/p;

The following one also speaks about the promotion of this issue

Http://www.cnblogs.com/openorz/archive/2011/11/14/2248992.html

The code is as follows:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring>using namespace std;const int maxn = 10000010;int prime[maxn],cnt;int ans[maxn];bool is[maxn];void init (int n) {    cnt=0;    Memset (Is,1,sizeof (IS));    memset (ans,0,sizeof (ans));    memset (prime,0,sizeof (Prime));    for (int i=2;i<=n;i++) {        if (Is[i]) {            prime[cnt++]=i;            for (int j=i+i;j<=n;j+=i)                is[j]=0;}}    } int fen (int n,int p) {    if (n==0) return 0;    Return Fen (n/p,p) +n/p;} int main () {    int n;    while (~SCANF ("%d", &n)) {        init (n);        for (int i=0;i<cnt;i++)            Ans[i]=fen (N,prime[i]);        for (int i=0;i<cnt;i++)            printf ("%d^%d\n", Prime[i],ans[i]);    }    return 0;}

n! decomposition of the element factor