BZOJ 1101 Zap(莫比乌斯反演)

http://www.lydsy.com/JudgeOnline/problem.php?id=1101

给定a,b,d，求有多少gcd(x,y)==d(1<=x<=a&&1<=y<=b)

思路:

Σgcd(x,y)==d  (1<=x<=a，1<=y<=b)

=

Σgcd(x,y)==1 (1<=x<=a/d,1<=y<=b/d)

令G(i)=num(i|gcd(x,y))=n/i*m/i

g(i)=num(i=gcd(x,y))

g(i)=ΣG(d)*u(d/i) (i|d)

则答案就是g(1)

g(1)=ΣG(i)*u(i) (1<=i<=min(n,m))

=Σ(n/i)*(m/i)*u(i)

因此做出u(i)的前缀和，这样就可以一起处理n/i和m/i相同的i

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 int mul[],mark[],sum[],p[];
 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;
 }
 void init(){
     mul[]=;
     for (int i=;i<=;i++){
         if (!mark[i]){
             p[++p[]]=i;
             mul[i]=-;
         }
         for (int j=;j<=p[]&&i*p[j]<=;j++){
             mark[p[j]*i]=;
             if (i%p[j]) mul[p[j]*i]=mul[i]*(-);
             else {
                 mul[p[j]*i]=;
                 break;
             }
         }
     }
     sum[]=;
     for (int i=;i<=;i++)
      sum[i]=sum[i-]+mul[i];
 }
 int cal(int a,int b){
     if (a>b) std::swap(a,b);
     int res=,n=a,m=b;
     for (int i=,j;i<=a;i=j+){
         j=std::min(n/(n/i),m/(m/i));
         res+=(a/i)*(b/i)*(sum[j]-sum[i-]);
     }
     return res;
 }
 int main(){
     int Q=read();
     init();
     while (Q--){
         int a=read(),b=read(),d=read();
         printf("%d\n",cal(a/d,b/d));
     }
 }