O

Given the value of N, you will have to find the value of G. The definition of G is given below:
G =
i

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

题解：

根据题意可以推得求的是1～n的与因子的最大公约数之和。
之前在51做过一道类似的，但是那是暴力枚举因子。
暴力枚举因子TLE了好几次，下次遇到因子一定要想到筛法啊。
T了几次之后，参考了别人。打表+前缀和优化

代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int MAX=4e6+10;typedef long long LL;LL phi[MAX];LL f[MAX];using namespace std;void get_eular(){     int i,j;     memset(phi,0,sizeof(phi));     phi[1]=1;     for(int i=2;i<MAX;i++)        if(!phi[i])     {         for(j=i;j<MAX;j+=i)         {             if(!phi[j]) phi[j]=j;             phi[j]=phi[j]/i*(i-1);         }     }     return ;}void init(){    memset(f,0,sizeof(f));    for(int i=1;i<MAX;i++)        for(int j=i*2;j<MAX;j+=i)//筛法枚举因子    {         f[j]+=phi[j/i]*i;    }    for(int i=3;i<MAX;i++)//前缀和优化    {        f[i]+=f[i-1];    }}int main(){    get_eular();    init();    LL N;    while((scanf("%lld",&N)!=EOF)&&N)    {      printf("%lld\n",f[N]);    }    return 0;}