[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[N],mark[N];int pri[N],phi[N],tot=0;long long sum[N];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;}