UVA 10214 Trees in a Wood (Eulerian function)

To enumerate each column, just find all y in g (x, y) = 1.

Because in 1 ~ x, if there are i coprime with x, then there are i in [1 + x, 2x], because each of the i coprime is added with x, and the common factor is still 1.

1≤y≤x There are phi (x)

x + 1≤y≤x also has phi (x)

2x + 1≤y≤x There are also phi (x)

.....

kx + 1≤y≤b .... direct statistics

#include<bits/stdc++.h>
using namespace std;
int phi[2005];
void euler_phi(int n){
	for(int i=2;i<=n;i++) phi[i]=0;
	phi[1]=1;
	for(int i=2;i<=n;i++)if(!phi[n]){
		for(int j=i;j<=n;j+=i){
			if(!phi[j]) 
			phi[j]=j;
			phi[j]=phi[j]/i*(i-1);
		}
	}	
	
} 

bool gcd(int a,int b){
	return b==0 ? a:gcd(b,a%b);
}

int main(){
	euler_phi(2000);
	int a,b;
	while(scanf("%d%d", &a, &b) == 2 && a){
		long long sum=(long long)4*a*b+(long long)2*(a+b);
		int trees=0;
		for(int x=1;x<=a;x++){
			int k=b/x;
			trees+=(long long)phi[x]*k;
			for(int y=k*x+1;y<=b;y++){
				if(gcd(x,y)) trees++;
			}
			
		}
		trees *= 4; trees += 4;
		double ans = (double)trees / sum;
		printf("%.7lf\n", ans);
	}
	
	
}