GCD - Extreme (II) UVA - 11426 数学

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;
}