BZOJ 2820 YY的GCD(莫比乌斯反演)

http://www.lydsy.com/JudgeOnline/problem.php?id=2820

 思路:

ans

=Σp(p为质数) Σgcd(x,y)==p (1<=x<=n,1<=y<=m)

=Σp(p为质数) Σgcd(x,y)==1 (1<=x<=n/p,1<=y<=m/p)

=Σp(p为质数) Σu(d) (d|gcd(x,y))  (1<=x<=n/p,1<=y<=m/p)

=Σp(p为质数) Σu(d) (d|x&&d|y)  (1<=x<=n/p,1<=y<=m/p)

=Σp(p为质数) Σd=1..n/p u(d)*floor(n/pd)*floor(m/pd)

k=pd

=Σp(p为质数) Σd=1..n/p u(k/p)*floor(n/k)*floor(m/k)

=Σk=1..n F(k)*floor(n/k)*floor(m/k)

可以预处理F数组

 1 #include<algorithm>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iostream>
 6 #define N 10000000
 7 #define ll long long
 8 int mark[N+10],p[N+10],mul[N+10],n,m;
 9 ll f[N+10];
10 int read(){
11     char ch=getchar();int t=0,f=1;
12     while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
13     while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
14     return t*f;
15 }
16 void init(){
17     mul[1]=1;
18     for (int i=2;i<=N;i++){
19         if (!mark[i]){
20             p[++p[0]]=i;
21             mul[i]=-1;
22         }
23         for (int j=1;j<=p[0]&&i*p[j]<=N;j++){
24             mark[p[j]*i]=1;
25             if (i%p[j]) mul[p[j]*i]=-mul[i];
26             else{
27                 mul[p[j]*i]=0;
28                 break;
29             }
30         }
31     }
32     for (int i=1;i<=p[0];i++){
33         int pri=p[i];
34         for (int j=1;j*pri<=N;j++)
35          f[pri*j]+=mul[j];
36     }
37     f[0]=0;
38     for (int i=1;i<=N;i++)
39       f[i]+=f[i-1];
40 }
41 int main(){
42     init();
43     int T=read();
44     while (T--){
45         ll ans=0;
46         n=read(),m=read();
47         if (n>m) std::swap(n,m);
48         for (int i=1,j;i<=n;i=j+1){
49             j=std::min(n/(n/i),m/(m/i));
50             ans+=(ll)(n/i)*(m/i)*(f[j]-f[i-1]);
51         }
52         printf("%lld\n",ans);
53     }
54 }

 

posted @ 2016-06-14 14:57  GFY  阅读(177)  评论(0)  编辑  收藏  举报