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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 4000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int phi[maxn];
void init() {
for (int i = 2; i <= maxn; i++)phi[i] = 0;
phi[1] = 1;
for (int i = 2; i <= maxn; i++) {
if (!phi[i]) {
for (int j = i; j <= maxn; j += i) {
if (!phi[j])phi[j] = j;
phi[j] = phi[j] / i * (i - 1);
}
}
}
}
ll sum[maxn], f[maxn];
int main() {
// ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
init();
for (int i = 1; i <= maxn; i++) {
for (int j = i * 2; j <= maxn; j += i)f[j] += i * phi[j / i];
}
sum[2] = f[2];
for (int j = 3; j <= maxn; j++)sum[j] = sum[j - 1] + f[j];
int n;
while (rdint(n) == 1&&n) {
printf("%lld\n", sum[n]);
}
return 0;
}