[HDU1695]GCD + [HAOI2011]Problem b + [POI2007]ZAP-Queries【莫比乌斯反演】

[HDU1695]GCD
[HAOI2011]Problem b
[POI2007]ZAP-Queries
令\[ans(n, m)=\sum_{i=1}^n\sum_{j=1}^m[GCD(i, j) == k]\]
\[=\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{k}\rfloor}[GCD(i, j) == 1]\]
令\[f(d)=\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{k}\rfloor}[GCD(i, j) == d]\]
\[g(x)=\sum_{x|d}f(d)\]
\[=\sum_{i=1}^{\lfloor\frac{n}{k}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{k}\rfloor}[x|GCD(i, j)]\]
\[=\sum_{i=1}^{\lfloor\frac{n}{xk}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{xk}\rfloor}[1|GCD(i, j)]\]
\[=\lfloor\frac{n}{xk}\rfloor\lfloor\frac{m}{xk}\rfloor\]
\[f(n)=\sum_{n|d}\mu(\frac{d}{n})g(d)\]
则\[ans(n, m)=f(1)=\sum_{d=1}^n\mu(d)g(d)\]
\[=\sum_{d=1}^n\mu(d)\lfloor\frac{n}{dk}\rfloor\lfloor\frac{m}{dk}\rfloor\]

[HDU1695]GCD
又\((x, y)\)和\((y,x)\)是等价的，要减去重复算的
假设\(b<d\),则最终\(ans=ans(b, d)-ans(b, b)/2\)

void init(){
    miu[1]=1;
    for(int i=2; i < N; i++) {
        if(!p[i]) p[++p[0]]=i, miu[i]=-1;
        for(int j=1; j <= p[0] && i*p[j] < N; j++){
            p[i*p[j]]=1; if(i%p[j] == 0) {miu[i*p[j]]=0; break;} else miu[i*p[j]]=-miu[i];
        }
    }
}
void solve(){
    init(); int T=read();
    for(int i=1; i <= T; i++){
        int a=read(), b=read(), c=read(), d=read(), k=read(); cout<<"Case "<<i<<": ";
        if(k == 0) {cout<<0<<endl; continue;} b/=k, d/=k;
        ll ans1=0, ans2=0, mn=min(b, d);
        for(int i=1; i <= mn; i++) 
            ans1+=1LL*miu[i]*(b/i)*(d/i), ans2+=1LL*miu[i]*(mn/i)*(mn/i);
        cout<<ans1-ans2/2<<endl;
    }
}

[HAOI2011]Problem b
这题要容斥一下\(ans=ans(b, d)-ans(a-1, d)-ans(b, c-1)+ans(a-1, c-1)\),
还要整除分块，否则会\(TLE\)

void init(){
    p[1]=miu[1]=1;
    for(int i=2; i < N; i++) {
        if(!p[i]) p[++p[0]]=i, miu[i]=-1;
        for(int j=1, x; j <= p[0] && (x=p[j]*i) < N; j++){ p[x]=1;
            if(i%p[j] == 0) {miu[x]=0; break;} miu[x]=-miu[i];
        }
    }
    for(int i=1; i < N; i++) miu[i]+=miu[i-1];
}
ll cal(ll m, ll n){ m/=k, n/=k;
    ll ans=0, mn=min(n, m); 
    for(int i=1, j; i <= mn; i=j+1){
        j=min(n/(n/i), m/(m/i));
        ans+=1LL*(miu[j]-miu[i-1])*(n/i)*(m/i);
    } 
    return ans;
}
void solve(){
    int T=read(); init();
    while(T--){
        ll a=read(), b=read(), c=read(), d=read(); k=read();
        printf("%lld\n", cal(b, d)-cal(a-1, d)-cal(b, c-1)+cal(a-1, c-1));
    }
}

[POI2007]ZAP-Queries 这题\(ans=ans(n, m)\)