#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

#include <set>

#include <map>

using namespace std;

typedef long long ll;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << (#x) << " = " << (x) << endl

#define error(x) (!(x)?puts("error"):0)

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)

const int N=10000005;

int p[N], cnt, np[N], mu[N], g[N], sum[N];

void init() {

	mu[1]=1;

	for2(i, 2, N) {

		if(!np[i]) p[++cnt]=i, mu[i]=-1, g[i]=1;

		for1(j, 1, cnt) {

			int t=p[j]*i; if(t>=N) break;

			np[t]=1;

			if(i%p[j]==0) { mu[t]=0; g[t]=mu[i]; break; }

			mu[t]=-mu[i]; g[t]=mu[i]-g[i];

		}

	}

	for2(i, 1, N) sum[i]=sum[i-1]+g[i];

}

int main() {

	init();

	int n=getint();

	ll ans=0;

	int pos;

	for(int i=1; i<=n; i=pos+1) {

		pos=min(n/(n/i), n/(n/i));

		ans+=(ll)(sum[pos]-sum[i-1])*(n/i)*(n/i);

	}

	printf("%lld\n", ans);

	return 0;

}

给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的
数对(x,y)有多少对.

一个整数N

如题

hint

对于样例(2,2),(2,4),(3,3),(4,2)

1<=N<=10^7

湖北省队互测