Given the value of
N
, you will have to nd the value of
G
. The de nition of
G
is given below:
G
=
i<N
∑
i
=1
j
N
∑
j
=
i
+1
GCD
(
i; j
)
Here
GCD
(
i; j
) means the greatest common divisor of integer
i
and integer
j
.
For those who have trouble understanding summation notation, the meaning of
G
is given in the
following code:
G=0;
for(i=1;i<N;i++)
for(j=i+1;j<=N;j++)
{
G+=gcd(i,j);
}
/*Here gcd() is a function that finds
the greatest common divisor of the two
input numbers*/
Input
The input le contains at most 100 lines of inputs. Each line contains an integer
N
(1
< N <
4000001).
The meaning of
N
is given in the problem statement. Input is terminated by a line containing a single
zero.
Output
For each line of input produce one line of output. This line contains the value of
G
for the corresponding
N
. The value of
G
will t in a 64-bit signed integer.
Sample Input
10
100
200000
0
Sample Output
67
13015
143295493160
设 f(n)=gcd(1,n)+gcd(2,n)+...+gcd(n-1,n)，则s(n)=f(2)+f(3)+...+f(n)；对于 f(n)=gcd(1,n)+...+gcd(n-1,n);设 g(n,i)= { gcd(x,n)=i 的个数 }，则 f(n)=Sum{ i*g(n,i) }；gcd(x,n)=i -->gcd(x/i,n/i)=1；那么满足条件的x/i有 phi(n/i)个；
#include
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#include
EPFL - Fighting
