UVA 11426

Given the value of N, you will have to find the value of G. The definition 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 file 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 fit in a 64-bit signed integer.
Sample Input
10
100
200000
0
Sample Output
67
13015
143295493160

题意：给出n，求∑（i!=j）   gcd(i,j)   （1<=i,j<=n）

题解：s(n)=s(n-1)+gcd(1,n)+gcd(2,n)+……+gcd(n-1,n);

设f(n)=gcd(1,n)+gcd(2,n)+……+gcd(n-1,n)。

gcd(x,n)=i是n的约数（x<n），按照这个约数进行分类。设满足gcd（x,n）=i的约束有g(n,i)个，则有f(n)=sum(i*g(n,i))。

而gcd(x,n)=i等价于gcd(x/i,n/i)=1，因此g(n,i)等价于phi(n/i).phi(x)为欧拉函数。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;

const int N=4000000+10;

ll phi[N+5] , f[N+5];
void phi_table() {
  for(int i = 2;i <= N; i++) phi[i] = 0;
  phi[1] = 1;
  for(int i = 2; i <= N; i++) {
    if(!phi[i]) {
        for(int j = i; j <= N; j += i) {
            if(!phi[j]) phi[j] = j;
            phi[j] = phi[j] / i * (i-1);
        }
    }
  }
}
ll s[N+5],n;
int main() {
    phi_table();
    for(int i = 1; i <= N; i++) {
        for(int j = i + i; j <= N; j += i) {
            f[j] += i * phi[j / i];
        }
    }
    for(int i = 1; i <= N; i++) s[i] = s[i-1] + f[i];
    while(~scanf("%lld",&n)) {
        if(!n) break;
        printf("%lld
", s[n]);
    }
    return 0;
}