BZOJ 3529 数表(莫比乌斯反演)

http://www.lydsy.com/JudgeOnline/problem.php?id=3529

思路:令F(i)为i的约数和，

1<=x<=n,1<=y<=m

G(i)为i|gcd(x,y)的个数

g(i)为i=gcd(x,y)的个数

G(i)=floor(n/i)*floor(m/i)

g(i)=Σu(d/i)*G(d) (i|d)

F(i)可以用线性筛O(n)处理

ans=(1<=i<=min(n,m)) Σg(i)*F(i)=Σ F(i)*ΣG(d)*u(d/i) (i|d)

=(1<=d<=min(n,m)) ΣG(d)* Σ(i|d) u(d/i)*F(i)

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define ll long long
 #define inf 0x7fffffff
 std::pair<int,int>f[];
 struct node{
     int n,m,id,a;
 }q[];
 int mx,v[],p[],mark[],mul[],ans[];
 int lowbit(int x){
     return x&(-x);
 }
 int read(){
     char ch=getchar();int t=,f=;
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 int Read(){
     char ch=getchar();ll t=,f=;
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 bool cmp(node a,node b){
     return a.a<b.a;
 }
 void add(int x,int y){
     for (int i=x;i<=mx;i+=lowbit(i)){
         v[i]+=y;
     }
 }
 int query(int x){
     int res=;
     if (x==) return ;
     for (int i=x;i;i-=lowbit(i)){
         res+=v[i];
     }
     return res;
 }
 void init(){
     mul[]=;
     for (int i=;i<=mx;i++){
         if (!mark[i]) {
             p[++p[]]=i;
             mul[i]=-;
         }
         for (int j=;j<=p[]&&i*p[j]<=mx;j++){
             mark[i*p[j]]=;
             if (i%p[j]) mul[i*p[j]]=mul[i]*(-);
             else {
                 mul[i*p[j]]=;
                 break;
             }
         }
     }
     for (int i=;i<=mx;i++)
      for (int j=i;j<=mx;j+=i)
       f[j].first+=i;
     for (int i=;i<=mx;i++) f[i].second=i;
 }
 void solve(int k){
     int id=q[k].id,n=q[k].n,m=q[k].m;
     for (int i=,j;i<=q[k].n;i=j+){
         j=std::min(n/(n/i),m/(m/i));
         ans[id]+=(n/i)*(m/i)*(query(j)-query(i-));
     }
 }
 int main(){
     int Q=read();
     for (int i=;i<=Q;i++){
         q[i].n=read(),q[i].m=read(),q[i].a=read();q[i].id=i;
         if (q[i].n>q[i].m) std::swap(q[i].n,q[i].m);
         mx=std::max(mx,q[i].n);
     }
     init();
     std::sort(q+,q++Q,cmp);
     std::sort(f+,f++mx);
     int now=;
     for (int i=;i<=Q;i++){
         while (now+<=mx&&f[now+].first<=q[i].a){
             now++;
             for (int j=f[now].second;j<=mx;j+=f[now].second){
                 add(j,f[now].first*mul[j/f[now].second]);
             }
         }
         solve(i);
     }
     for (int i=;i<=Q;i++)
      printf("%d\n",ans[i]&inf);
 }