Gcd
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
题意:
略
分析:
和这道题完全一样
只不过不需要优化直接两重for循环枚举素数即可
code:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1e7+10;
typedef long long LL;
//LL F[MAXN],f[MAXN];
LL pri[MAXN],pri_num;
LL mu[MAXN];//莫比乌斯函数值
bool vis[MAXN];
void mobius(int N) //筛法求莫比乌斯函数
{
pri_num = 0;//素数个数
memset(vis, false, sizeof(vis));
vis[1] = true;
mu[1] = 1;
for(int i = 2; i <=N; i++)
{
if(!vis[i])
{
pri[pri_num++] = i;
mu[i] = -1;
}
for(int j=0; j<pri_num && i*pri[j]<N ; j++)
{
vis[i*pri[j]]=true;//标记非素数
//eg:i=3,i%2,mu[3*2]=-mu[3]=1;----;i=6,i%5,mu[6*5]=-mu[6]=-1;
if(i%pri[j])mu[i*pri[j]] = -mu[i];
else
{
mu[i*pri[j]] = 0;
break;
}
}
}
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
LL n;
mobius(10000000);
while(~scanf("%lld",&n))
{
LL ans = (LL)0;
for(LL i=0; pri[i]<=n; i++)
{
for(LL j=1; j<=n/pri[i]; j++)
ans+=(LL)(mu[j]*((n/pri[i])/j)*((n/pri[i])/j));
}
printf("%lld\n",ans);
}
return 0;
}