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uva10214 Trees in a Wood.

类似 https://www.cnblogs.com/towboa/p/17303216.html ,

不过给的是n ,m (n<=2000)

枚举 i (1<=i<=n) ，考虑有多少 j （1<=j<=m) gcd__(i,j)==0

然后分段考虑

(gcd(x,y) = gcd(x,y-x) )

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std ; 
const int N =3e5+4;
  
 const int M=1e6;
 
 int vis[M+4],P[M+4],cnt;
 int fi[M+4];
 int sum[N] ;
  
 int gcd(int x,int y){
    return y==0?x: gcd(y,x%y);
 }
 void shai(int top){
     cnt=0;
     fi[1]=1;
     for(int i=2;i<=top;i++){
         if(vis[i]==0){
             P[++cnt]=i; 
             fi[i]=i-1;
         }
         for(int j=1;j<=cnt&&i*P[j]<=top;j++){
             vis[i*P[j]]=1;
             if(i%P[j]==0){
                 fi[i*P[j]]=fi[i]*P[j];
                 break;
             }
            else
            fi[i*P[j]]=fi[i]*(P[j]-1);
         }
     }
 }
 signed main(){
    int x,y;
     
    shai(2002);
    int a,b; long long ans;
    while(scanf("%d%d",&a,&b)!=EOF){
        if(!a) return 0;
        ans=0;
        ans+=b+1;
        for(int i=2;i<=a;i++) 
        {
            ans+=1ll*b/i*fi[i];
            for(int j=b/i*i+1;j<=b;j++) ans+=__gcd(i,j)==1;
        }
        ans<<=2;
        printf("%.7lf\n",1.0*ans/(1ll*a*b*4+2*(a+b)));
    }
 }