[bzoj2818]GCD 欧拉函数线筛

2818: Gcd

Time Limit: 10 Sec  Memory Limit:
256 MB
Submit: 5186  Solved: 2327

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Description

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.

Input

一个整数N

Output

如题

Sample Input

4

Sample Output

4

HINT

hint

对于样例(2,2),(2,4),(3,3),(4,2)

1<=N<=10^7

Source

枚举1到n中每个素数p

分别用每个数除p,即求互质对的个数

用欧拉函数*2-1即可

#include<iostream>
#include<cstdio>
using namespace std;
const int N = 10000005;
bool isnot
,mark
;
int pri
,phi
,tot=0;
long long sum
;
void init( int n ){
isnot[1] = true; phi[1] = 1;
for( int i = 2; i <= n; i++ ){
if( !isnot[i] ){
pri[++tot] = i;
phi[i] = i - 1;
}
for( int t = 1; t <= tot; t++ ){
int j = pri[t]*i; if( j > n ) break;
isnot[j] = true;
if( i % pri[t] ) phi[j] = phi[i] * phi[pri[t]];
else {
phi[j] = pri[t] * phi[i];
break;
}
}
}
for( int i = 2; i <= n; i++ ) sum[i] = sum[i-1] + phi[i];
}
int n;
long long ans=0;
int main(){
scanf("%d", &n);
init(n);
for( int i = 1; i <= n; i++ ) sum[i] = sum[i-1] + phi[i];
for( int i = 1; i <= tot; i++ ) ans += sum[n/pri[i]]*2-1;
printf("%lld",ans);
return 0;
}