[Luogu2257]YY的GCD
求\(\sum_{i=1}^n\sum_{j=1}^m[(x, y)\)為質數\(]\)
\(T \le {10}^4, 1\le n, m \le {10}^7\)
假設\(p\)為\(M\)以內的質數
\[\begin{aligned}ans&=\sum_{p\in prime}\sum_{i=1}^n\sum_{j=1}^m[(i, j) == p]\&=\sum_{p\in prime}\sum_{x=1}^{\lfloor\frac{n}{p}\rfloor}\mu(x)\lfloor\frac{n}{xp}\rfloor\lfloor\frac{m}{xp}\rfloor\end{aligned}\]

void init(){
    miu[1]=1;
    for(int i=2; i < Maxn; i++){
        if(!p[i]) p[++ptot]=i, miu[i]=-1;
        for(int j=1, x; j <= ptot && (x=p[j]*i) < Maxn; j++){
            p[x]=1; if(i%p[j] == 0) break; miu[x]=-miu[i];
        }
    }
    for(int i=1; i <= ptot; i++) for(int j=p[i]; j < Maxn; j+=p[i]) G[j]+=miu[j/p[i]];
    for(int i=1; i < Maxn; i++) G[i]+=G[i-1];
}
void solve(){
    init(); int T=read();
    while(T--){
        n=read(), m=read(); ll ans=0; if(n > m) swap(n, m);
        for(int l=1, r=0; r < n; l=r+1){
            r=min(n/(n/l), m/(m/l));
            ans+=1ll*(G[r]-G[l-1])*(n/l)*(m/l);
        }
        printf("%lld\n", ans);
    }
}
 

YY的GCD【莫比烏斯反演】