Gcd

Time Limit: 10 Sec Memory Limit: 256 MB

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

1<=N<=10^7

Solution

直接莫比乌斯反演即可。

img

然后对于这个式子，我们下界分块一下即可。

Code

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#include<bits/stdc++.h>
using namespace std;
typedef long long s64;

const int ONE = 1e7+5;

int T;
int n,m;
bool isp[ONE];
int prime[664580],p_num;
int miu[ONE],sum_miu[ONE];
s64 Ans;

int get()
{
    int res=1,Q=1;  char c;
    while( (c=getchar())<48 || c>57)
        if(c=='-')Q=-1;
    if(Q) res=c-48;
    while((c=getchar())>=48 && c<=57)
        res=res*10+c-48;
    return res*Q;
}

void Getmiu(int MaxN)
{
    miu[1] = 1;
    for(int i=2; i<=MaxN; i++)
    {
        if(!isp[i])
            isp[i] = 1, prime[++p_num] = i, miu[i] = -1;
        for(int j=1; j<=p_num, i*prime[j]<=MaxN; j++)
        {
            isp[i * prime[j]] = 1;
            if(i % prime[j] == 0)
            {
                miu[i * prime[j]] = 0;
                break;
            }
            miu[i * prime[j]] = -miu[i];
        }
        miu[i] += miu[i-1];
    }
}

int main()
{
    n=get();
    Getmiu(n);
    for(int d=1; d<=p_num; d++)
    {
        if(prime[d] > n) break;
        int N = n/prime[d];
        for(int i=1,j=0; i<=N; i=j+1)
        {
            j = min(N, N/(N/i));
            Ans += (s64)(N/i) * (N/i) * (miu[j] - miu[i-1]);
        }
    }

    printf("%lld",Ans);
}