莫比乌斯反演μ函数模板

参考学习：

https://blog.csdn.net/HowardEmily/article/details/78199037

https://blog.csdn.net/codeswarrior/article/details/81541972

O( n )筛法函数:

void init(){
    memset(prime,0,sizeof(prime));
    memset(mu,0,sizeof(mu));
    memset(vis,0,sizeof(vis));
    mu[1] = 1;
    cnt = 0;
    for(int i = 2; i < maxn; i++){
        if(!vis[i]){
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && i * prime[j] < maxn; j++){
            vis[i*prime[j]] = 1;
            if(i % prime[j]) mu[i*prime[j]] = -mu[i];
            else{
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}

HDU 1695 GCD 反演经典入门：

http://acm.hdu.edu.cn/showproblem.php?pid=1695

Ac code：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
typedef long long ll;
const int maxn = 1e5+7;
bool vis[maxn];
int prime[maxn],mu[maxn];
int cnt;
void init(){
    memset(prime,0,sizeof(prime));
    memset(mu,0,sizeof(mu));
    memset(vis,0,sizeof(vis));
    mu[1] = 1;
    cnt = 0;
    for(int i = 2; i < maxn; i++){
        if(!vis[i]){
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && i * prime[j] < maxn; j++){
            vis[i*prime[j]] = 1;
            if(i % prime[j]) mu[i*prime[j]] = -mu[i];
            else{
                mu[i*prime[j]] = 0;
                break;
            }
        }
    }
}
int main(){
    init();
    int T_case;
    int a, b, c, d, k;
    LL ans1 = 0, ans2 = 0;
    scanf("%d", &T_case);
    for(int Case = 1; Case <= T_case; Case++){
        ans1 = 0; ans2 = 0;
        printf("Case %d: ", Case);
        scanf("%d %d %d %d %d", &a, &b, &c, &d, &k);
        if(b < k || d < k || k == 0){
            puts("0");
            continue;
        }
        b = b/k;
        d = d/k;
        int len = min(b, d);
        for(int i = 1; i <= len; i++){
            ans1 += (LL)mu[i]*(b/i)*(d/i);
        }
        for(int i = 1; i <= len; i++){
            ans2 += (LL)mu[i]*(len/i)*(len/i);
        }
        printf("%lld
", ans1-ans2/2);
    }
    return 0;
}