BZOJ 2301 Problem B(莫比乌斯反演)

http://www.lydsy.com/JudgeOnline/problem.php?id=2301

题意:给a,b,c,d,k,求gcd(x,y)==k的个数(a<=x<=b,c<=y<=d)

思路:假设F(a,b)代表gcd(x,y)==k 的个数(1<=x<=a,1<=y<=b)

那么这是满足区间加减的

ans=F(b,d)-F(b,c)-F(a,d)+F(a,c)

剩下的就和Zap一样了

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