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

 1 #include<algorithm>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iostream>
 6 int mul[200005],mark[200005],sum[200005],p[200005];
 7 int read(){
 8     char ch=getchar();int t=0,f=1;
 9     while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
10     while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
11     return t*f;
12 }
13 void init(){
14     mul[1]=1;
15     for (int i=2;i<=50000;i++){
16         if (!mark[i]){
17             p[++p[0]]=i;
18             mul[i]=-1;
19         }
20         for (int j=1;j<=p[0]&&i*p[j]<=50000;j++){
21             mark[p[j]*i]=1;
22             if (i%p[j]) mul[p[j]*i]=mul[i]*(-1);
23             else {
24                 mul[p[j]*i]=0;
25                 break;
26             }
27         }
28     }
29     sum[0]=0;
30     for (int i=1;i<=50000;i++)
31      sum[i]=sum[i-1]+mul[i];
32 }
33 int cal(int a,int b){
34     if (a>b) std::swap(a,b);
35     int res=0,n=a,m=b;
36     for (int i=1,j;i<=a;i=j+1){
37         j=std::min(n/(n/i),m/(m/i));
38         res+=(a/i)*(b/i)*(sum[j]-sum[i-1]);
39     }
40     return res;
41 }
42 int main(){
43     int Q=read();
44     init();
45     while (Q--){
46         int a=read(),b=read(),d=read();
47         printf("%d\n",cal(a/d,b/d));
48     }
49 }

 

posted @ 2016-06-14 10:59  GFY  阅读(370)  评论(0)  编辑  收藏  举报