GCD BZOJ2818 [省队互测] 数学

题目描述

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.

输入输出格式

输入格式：

一个整数N

输出格式：

答案

输入输出样例

输入样例#1： 复制

4

输出样例#1： 复制

4

说明

对于样例(2,2),(2,4),(3,3),(4,2)

1<=N<=10^7

来源：bzoj2818

本题数据为洛谷自造数据，使用CYaRon耗时5分钟完成数据制作。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 2000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

bool vis[10000002];
int phi[10000002];
ll sum[10000002 >> 1];
int pri[10000002];
int tot;
int N;
void init() {
	phi[1] = 1;
	for (int i = 2; i <= 10000000; i++) {
		if (!vis[i]) {
			pri[++tot] = i; phi[i] = i - 1;
		}
		for (int j = 1; j <= tot && i*pri[j] <= 10000000; j++) {
			vis[i*pri[j]] = true;
			phi[i*pri[j]] = phi[i] * phi[pri[j]];
			if (i%pri[j] == 0) {
				phi[i*pri[j]] = phi[i] * pri[j];
				break;
			}
		}
	}
}

int main()
{
	//	ios::sync_with_stdio(0);
	N = rd();
	init();
	for (int i = 1; i <= 10000000 / 2; i++)sum[i] = 1ll*sum[i - 1] + 1ll*phi[i];
	ll ans = 0;
	for (int i = 1; i <= tot; i++) {
		if (pri[i] > N)break;
		ans += 1ll * 2 * sum[N / pri[i]] - 1;
	}
	cout << (ll)ans << endl;
	return 0;
}