ch3101 阶乘分解（数学+思维）

• 直接分解显然会炸
• 于是我们可以先筛出小于N的所有质数
• 对于每个质因子p，1~N中包含一个p的数有[N/p]个，包含两个的有[N/p^2]个…
• 由于后一包含前一，所以总共有个质因子p
• O(NlogN)

//#include<bits/stdc++.h>
#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include  <cassert>
#include   <cstdio>
#include   <vector>
#include   <string>
#include    <cmath>
#include    <queue>
#include    <stack>
#include      <set>
#include      <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
ll gcd(ll a, ll b) {
	return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B) {
	(A<B)&&(A=B);//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B) {
	(A>B)&&(A=B);//if(B<A)A=B;
}
template <class T>
inline bool scan_d(T &ret) {
	char c;
	int sgn;
	if(c=getchar(),c==EOF) return 0; //EOF
	while(c!='?'&&(c<'0'||c>'9')) c=getchar();
	sgn=(c=='?')?-1:1;
	ret=(c=='?')?0:(c-'0');
	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
	ret*=sgn;
	return 1;
}
inline void outt(int x) {
	if(x>9) outt(x/10);
	putchar(x%10+'0');
}
const int maxn=1e6+10;
int vis[maxn];
struct node{
	ll prime;
	ll cnt;
}P[maxn];
int cnt;
void init(int n){
	vis[1]=1;
	cnt=0;
	rep(i,1,n){
		if(!vis[i]){
			P[++cnt].prime=i;
			P[cnt].cnt=0;
			for(int j=i*2;j<=n;j+=i){
				if(!vis[j])vis[j]=1;
			}
		}
	}
}
int main(){
	
	int n;
	scanf("%d", &n);
	init(n);
	rep(i,1,cnt){
		ll base=P[i].prime;
		ll num=base;
		while(num<=n){
			P[i].cnt+=n/num;
			num*=base;
		}
	}
	rep(i,1,cnt){
		printf("%lld %lld",P[i].prime,P[i].cnt);
		if(i!=cnt)puts("");
	}
	return 0;
}