题目链接：https://ac.nowcoder.com/acm/contest/9536/S

题解：

一道反演的比较基础的题，进行一下比较简单的转换就行了。不过需要对反演的知识有一定的了解，不懂的可以参考：我也不知道什么是"莫比乌斯反演"和"杜教筛"

 推导过程：

 

代码实现：

#include <bits/stdc++.h>
#define PI atan(1.0)*4
#define rp(i,s,t) for (int i = (s); i <= (t); i++)
#define RP(i,t,s) for (int i = (t); i >= (s); i--)
#define sc(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset(a,b,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define m_p make_pair
#define p_b push_back
#define ins insert
#define era erase
#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define dg if(debug)
#define outval(a) cout << "Debuging...|" << #a << ": " << a << "\n";
using namespace std;
int debug = 0;
ll gcd(ll a,ll b){
	return b?gcd(b,a%b):a;
}
ll lcm(ll a,ll b){
	return a/gcd(a,b)*b;
}
inline int read(){
	int s=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-') f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		s=s*10+ch-'0';
		ch=getchar();
	}
	return s*f;
}
const int N = 1e6+7;
int n;
ll phi[N],mu[N];
int prime[N];
int flag[N];
ll F[N];
ll G[N];
ll Num[N];
int num=0;
int f(int x){
	int ans=0;
	while(x) ans+=x%10,x/=10;
	return ans;
}
int g(int x){
	int ans=1;
	while(x) ans*=x%10,x/=10;
	return ans;
}
void init()
{
	phi[1]=1;
	mu[1]=1;
	F[1]=G[1]=1;
	for (int i=2;i<=n;i++){
		F[i]=f(i);
		G[i]=g(i);
		if (flag[i]==0)//这代表i是质数 
		{
			prime[++num]=i;
			phi[i]=i-1;
			mu[i]=-1;
		}
		for (int j=1;j<=num&&prime[j]*i<=n;j++){
			flag[i*prime[j]]=1;
			if (i%prime[j]==0){
				phi[i*prime[j]]=phi[i]*prime[j];
				mu[i*prime[j]]=0;
				break;
			}
			phi[i*prime[j]]=phi[i]*(prime[j]-1),mu[i*prime[j]]=-mu[i];
		}
	}
	rp(i,1,n) for(int j=i;j<=n;j+=i) Num[j]+=mu[i]*(n/i-j/i+1);
}
void solve(){
	n=read();
	init();
	ll ans=0;
	rp(i,1,n){
		ans+=phi[i]*F[i];
		ans+=Num[i]*G[i];
	}
	cout<<ans<<endl;
}
int main(){
	//ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#ifdef ONLINE_JUDGE
#else
	freopen("in.txt", "r", stdin);
	//debug = 1;
#endif
	time_t beg, end;
	//if(debug) beg = clock();
	// int T=read();
	// while(T--)
	solve();

	/*
	if(debug) {
		end = clock();
		printf("time:%.2fs\n", 1.0 * (end - beg) / CLOCKS_PER_SEC);
	}
	*/
	return 0;
}