YY的GCD【莫比乌斯反演】

[Luogu2257]YY的GCD
求(sum_{i=1}^nsum_{j=1}^m[(x, y))为质数(])
(T le {10}^4, 1le n, m le {10}^7)
假设(p)为(M)以内的质数

[egin{aligned}ans&=sum_{pin prime}sum_{i=1}^nsum_{j=1}^m[(i, j) == p]\ &=sum_{pin prime}sum_{x=1}^{lfloorfrac{n}{p} floor}mu(x)lfloorfrac{n}{xp} floorlfloorfrac{m}{xp} floorend{aligned}]

令(T=xp)则

[egin{aligned}ans&=sum_{T=1}^nsum_{pin prime ;&&; p|T}{mu(frac{T}{p})lfloorfrac{n}{T} floorlfloorfrac{m}{T} floor}\ &=sum_{T=1}^nlfloorfrac{n}{T} floorlfloorfrac{m}{T} floorsum_{pin prime ;&&; p|T}mu(frac{T}{q}) end{aligned} ]

只要筛出(G(T)=sum_{pin prime ;&&; p|T}mu(frac{T}{q}))就可以(O(sqrt{N}))算出

[ans=sum_{T=1}^nlfloorfrac{n}{T} floorlfloorfrac{m}{T} floor G(T) ]

这题除了(ans)都不能开(long;long)，不然会(TLE)

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
", ans);
	}
}