BZOJ 2818-Gcd（莫比乌斯反演）

2818: Gcd

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 5969  Solved: 2648
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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

题意：求有多少对（x,y）满足gcd（x，y）为质数

题解：我们设f(n)为gcd（x，y）=n的对数，F(n)为gcd（x,y）为n的倍数的对数

然后我们可以利用莫比乌斯反演定理，将mu函数带入进来。

#include<map>     #include<stack>            #include<queue>            #include<vector>    #include<string>  #include<math.h>            #include<stdio.h>            #include<iostream>            #include<string.h>            #include<stdlib.h>    #include<algorithm>   #include<functional>    using namespace std;typedef long long  ll;#define inf  1000000000       #define mod 1000000007             #define maxn  10000005  #define lowbit(x) (x&-x)            #define eps 1e-9  int a[maxn] = { 1,1 }, b[maxn/10], mu[maxn], cnt;void init(){ll i, j;mu[1] = 1;for (i = 2;i < maxn;i++){if (a[i] == 0)b[++cnt] = i, mu[i] = -1;for (j = 1;j <= cnt && b[j] * i < maxn;j++){a[b[j] * i] = 1;if (i % b[j] == 0){mu[b[j] * i] = 0;break;}elsemu[b[j] * i] = -mu[i];}}}int main(void){init();ll i, j, n, ans = 0;scanf("%lld", &n);for (i = 1;i <= cnt && b[i] <= n;i++){ll sum = 0;for (j = b[i];j <= n;j += b[i])sum += (ll)mu[j / b[i]] * (n / j)*(n / j);ans += sum;}printf("%lld\n", ans);return 0;}