[国家集训队]Crash的数字表格【莫比乌斯反演】

[luoguP1829][国家集训队]Crash的数字表格 / JZPTAB
求$$\sum_{i=1}^n\sum_{j=1}mLCM(i, j); mod; 20101009$$,
\(n, m\le {10}^7\)
以下\(n \le m\)

\[ans=\sum_{i=1}^n\sum_{j=1}^mLCM(i, j) \]

\[=\sum_{i=1}^n\sum_{j=1}^m\frac{ij}{GCD(i, j)} \]

\[=\sum_{d=1}^n\sum_{i=1}^n\sum_{j=1}^m\frac{ij[GCD(i, j) == d]}{d} \]

\[=\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}ij[GCD(i, j) == 1] \]

令$$f(d)=\sum_{i=1}^x\sum_{j=1}yij[GCD(i,j) == d]$$

\[g(d)=\sum_{d|n}f(n) \]

则$$g(d)=\sum_{i=1}^x\sum_{j=1}yij[d|GCD(i,j)]$$

\[=d^2\sum_{i=1}^{\lfloor\frac{x}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{y}{d}\rfloor}ij[1|GCD(i,j)] \]

令\(sum(x)=\sum_{i=1}^xi\)
则$$g(d)=d^2\sum_{i=1}\rfloor}i*sum(\lfloor\frac{y}{d}\rfloor)$$

\[=d^2*sum(\lfloor\frac{x}{d}\rfloor)*sum(\lfloor\frac{y}{d}\rfloor) \]

则$$f(n)=\sum_{n|d}\mu(\frac{d}{n})g(d)$$

\[f(1)=\sum_{d=1}^x\mu(d)g(d) \]

\[=\sum_{d=1}^x\mu(d)d^2*sum(\lfloor\frac{x}{d}\rfloor)*sum(\lfloor\frac{y}{d}\rfloor) \]

\[ans=\sum_{d=1}^nd\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\mu(x)x^2*sum(\lfloor\frac{n}{xd}\rfloor)*sum(\lfloor\frac{m}{xd}\rfloor) \]

好像到这样就可以了？

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