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[200005],mark[200005],sum[200005],p[200005];
int read(){
    char ch=getchar();int t=0,f=1;
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
void init(){
    mul[1]=1;
    for (int i=2;i<=50000;i++){
        if (!mark[i]){
            p[++p[0]]=i;
            mul[i]=-1;
        }
        for (int j=1;j<=p[0]&&i*p[j]<=50000;j++){
            mark[p[j]*i]=1;
            if (i%p[j]) mul[p[j]*i]=mul[i]*(-1);
            else {
                mul[p[j]*i]=0;
                break;
            }
        }
    }
    sum[0]=0;
    for (int i=1;i<=50000;i++)
     sum[i]=sum[i-1]+mul[i];
}
int cal(int a,int b){
    if (a>b) std::swap(a,b);
    int res=0,n=a,m=b;
    for (int i=1,j;i<=a;i=j+1){
        j=std::min(n/(n/i),m/(m/i));
        res+=(a/i)*(b/i)*(sum[j]-sum[i-1]);
    }
    return res;
}
int main(){
    int Q=read();
    init();
    while (Q--){
        int a=read(),b=read(),d=read();
        printf("%d
",cal(a/d,b/d));
    }
}