题解：艾米利亚的求助

题目:http://cojs.tk/cogs/problem/problem.php?pid=2431

题解:

   其实这道题真心简单，不要被数据范围迷惑。

对于F(x)函数，我们把x分解为p1^a1*p2^a2*……*pn^an

F(x)=(a1+1)*(a2+1)*……*(an+1); 

         直接枚举N的因子，已知N的因子个数是logN个，所以后面的就直接暴力就可以了，效率是log(N)*，10^15可以过掉

#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
const int MAXN=10000005;
ll N,Ans,tot;
ll ans_1[MAXN],ans_2[MAXN],Tmp[MAXN];
ll Phi(ll x){
	ll res=x,k=sqrt(x);
	for(int i=2;i<=k;++i){
		if(x%i==0){
			res-=res/i;
			while(x%i==0)x/=i;
		}
	}
	if(x>1)res-=res/x;
	return res;
}
 
ll F(ll x){
	Tmp[0]=0;
	int k=sqrt(x);
	ll ans=1;
	for(int i=2;i<=k;i++){
		while(x%i==0){
			Tmp[++Tmp[0]]=i;
			x/=i;
		}
	}
	if(x>1)Tmp[++Tmp[0]]=x;
	if(!Tmp[0])return 0;
	for(int i=1,j=i;i<=Tmp[0];i++){
		j=i;
		while(Tmp[j] == Tmp[j+1] && j<Tmp[0])j++;
		ans*=(j-i+2); i=j;
	}
	return ans;
}
 
int main(){
	freopen("aimiliyadehelp.in","r",stdin);
	freopen("aimiliyadehelp.out","w",stdout);
	scanf("%lld",&N);
	int k=sqrt(N);
	for(int i=1;i<=k;++i){
		if(N%i==0){
			ans_1[++ans_1[0]]=i;
			if(i*i!=N)ans_1[++ans_1[0]]=N/i;
		}
	}
	for(int i=1;i<=ans_1[0];++i)
		Ans+=F(ans_1[i])*Phi(N/ans_1[i]);
	printf("%lld
",Ans);
	fclose(stdin);
	fclose(stdout);
	return 0;
}