BZOJ2818 GCD 【莫比乌斯反演】

2818: Gcd

Time Limit: 10 Sec Memory Limit: 256 MB

Submit: 6826 Solved: 3013

[Submit][Status][Discuss]

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

题解

一般gcd一堆求和都是莫比乌斯

我们设f(n)表示gcd等于n的对数

我们设F(n)表示n|gcd的对数

则有

F(n)=⌊Nn⌋2

f(n)=∑n|dμ(dn)F(d)

=∑n|dμ(dn)⌊Nn⌋2

=∑Ni=1μ(i)⌊Ni∗n⌋2

ans=∑Np∈prime∑Ni=1μ(i)⌊Ni∗p⌋2

=∑NT=1⌊NT⌋2∗∑Np|Tμ(Tp)

至此我们可以枚举T，之后计算后边的和式就好了

其实后边的和式可以预处理得到，我直接算也能过

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)
using namespace std;
const int maxn = 10000005,maxm = 100005,INF = 1000000000;
bitset<maxn> isn;
int prime[maxn],primei,miu[maxn],N;
void init(){
    miu[1] = 1;
    for (int i = 2; i <= N; i++){
        if (!isn[i]) prime[++primei] = i,miu[i] = -1;
        for (int j = 1; j <= primei && i * prime[j] <= N; j++){
            isn[i * prime[j]] = true;
            if (i % prime[j] == 0) {miu[i * prime[j]] = 0;break;}
            miu[i * prime[j]] = -miu[i];
        }
    }
}
int main(){
    cin>>N;
    init();
    LL ans = 0;
    for (int i = 1; i <= primei; i++){
        for (int j = 1; j <= N / prime[i]; j++)
            ans += (LL) miu[j] * (N / prime[i] / j) * (N / prime[i] / j);
    }
    cout<<ans<<endl;
    return 0;
}