Trees in a Wood. UVA - 10214 欧拉函数表

紫书p339

枚举a，每个的贡献为b/i*phi[i]，因为gcd(kx+y,x)=gcd(x,y),因为gcd(x,y)=gcd(y,x%y)

#include <iostream>
#include <cstdio>
#include <ctime>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <set>
#include <queue>
using namespace std;
typedef unsigned long long ll;
const int INF=1e9+100;
const int mod=1e9+7;

int phi[2005];
void phi_table(int n){
memset(phi,0,sizeof(phi));
phi[1]=1;
for(int i=2;i<=n;i++){
if(!phi[i]){
for(int j=i;j<=n;j+=i){
if(!phi[j]) phi[j]=j;
phi[j]=phi[j]/i*(i-1);
}
}
}
}

ll gcd(ll x,ll y){
return y?gcd(y,x%y):x;
}

int main(){
//freopen("out.txt","w",stdout);
int a,b;
phi_table(2001);
while(scanf("%d %d",&a,&b)&&a+b){
ll ans=0;
for(int i=1;i<=a;i++){
int t=b/i;
ans+=1LL*t*phi[i];
for(int j=t*i+1;j<=b;j++){
if(gcd(j,i)==1) ans++;
}
}
ans=1LL*ans*4+4;
double ans1=1.0*ans/((1LL*2*a+1)*(2*b+1)-1);
printf("%.7f\n",ans1 );
}
return 0;
}