莫比乌斯反演模板

#include <bits/stdc++.h>

using namespace std;
const int mx = 1e5+5;
typedef long long ll;

bool vis[mx];
int sum[mx], prim[mx], mu[mx], cnt = 0;

void get_mu(){
    mu[1] = 1;
    for(int i = 2; i< mx; i++){
        if(!vis[i]) {
            prim[++cnt] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= cnt && prim[j]*i < mx; j++) {
            vis[prim[j]*i] = 1;
            if(i % prim[j] == 0) break;
            else mu[i*prim[j]] = -mu[i];
        }
    }
    sum[0] = 0;
    for (int i = 0; i < mx; i++) sum[i] = sum[i-1]+ mu[i];
}


int main() {
    get_mu();
    int n, m;
    scanf("%d%d", &n, &m);

    int len = min(n, m);
    int last;

    ll ans = 0;
    /*利用n/i连续区间数值相同的性质进行分块计算，可把复杂度降为sqrt(n) */
    for (int i = 1; i <= len; i=last+1) {
        last = min(n/(n/i), m/(m/i));
        ans += 1LL * (sum[last] - sum[i-1]) * (n/i) * (m/i);
    }
    printf("%lld
", ans);

}