uva10214 Trees in a Wood.

The saying “You can’t see the wood for the trees” is not only a

cliche, but is also incorrect. The real problem is that you can’t see

the trees for the wood. If you stand in the middle of a wood, the

trees tend to obscure each other and the number of distinct trees you

can actually see is quite small. This is especially true if the trees

are planted in rows and columns, because they tend to line up. The

purpose of this problem is to find how many distinct trees one can see

if one were standing on the position of a tree in the middle of the

wood. For a mathematically more precise description we assume that you

are situated at the origin of a coordinate system in the plane. Trees

are planted at all positions (x,y) ∈ ZZ×ZZ{(0,0)}, with |x|≤ a and

|y|≤ b. A tree at position (x,y) can be seen from the origin if there

are no other trees on the straight line from (0,0) to (x,y). Find the

number K of all the trees in the wood that can be seen from the origin

and the number N of all the trees to compute the fraction K N . Hint:

The Euler phi function φ(n) is defined to be the number of integers m

in the range 1 ≤ m ≤ n relatively prime to n: φ(n) = #{m | 1 ≤ m ≤ n

and gcd(m,n) = 1} (gcd = greatest common divisor) Instead of counting

(an adequate method for small n!) you could as well use the following

identity: φ(n) = n ∏ p∈P,p|n(1− 1 p), P = {p ∈ IN|p prime} Hint:

Remember that gcd(m,n) = gcd(m + n,n) = gcd(m + 2n,n) = gcd(m + in,i)

You might be surprised that the fraction K N converges to 6 π2 ≈

0.607927 for an infinitely largewood. Input Each scenario consists of a line with the two numbers a and b (separated by a white space), with 1

≤ a ≤ 2000 and 1 ≤ b ≤ 2000000. Input is terminated by a line with two

zeros. Output For each scenario print a line containing the fraction

with a precision of 7 digits after the decimal point. Error less than

2e-7 or 2∗10−7 will be tolerated.

求出所有互质的数对(x,y)(1<=x<=a,1<=y<=b)的个数ans，那么答案就是(4 * ans+4)/((2 * a+1) * (2 * b+1)-1)。

因为a比较小，b比较大，所以枚举i=1..a，计算1..b中与i互质的数。

不难发现，在区间[1..i],[i+1..2*i],…中各有phi(i)个满足条件的数，最后一段长度不正好为i的O(i)暴力即可。

#include<cstdio>
#include<cstring>
#define LL long long
int phi[2010],gcd[2010][2010];
int find(int a,int b)
{
if (b==0) return a;
if (gcd[a][b]) return gcd[a][b];
return gcd[a][b]=find(b,a%b);
}
int main()
{
int i,j,k,p,q,a,b;
LL x,y,z,ans;
phi[1]=1;
for (i=2;i<=2000;i++)
if (!phi[i])
for (j=i;j<=2000;j+=i)
{
if (!phi[j]) phi[j]=j;
phi[j]=phi[j]/i*(i-1);
}
for (i=1;i<=2000;i++)
for (j=1;j<=2000;j++)
gcd[i][j]=find(i,j);
while (scanf("%d%d",&a,&b)&&a)
{
ans=0;
for (i=1;i<=a;i++)
{
ans+=(LL)phi[i]*(b/i);
for (j=1;j<=b%i;j++)
if (gcd[i][j]==1)
ans++;
}
printf("%.7f\n",((double)ans*4+4)/((2*(double)a+1)*(2*b+1)-1));
}
}