fzu1969 GCD Extreme 类似于uva10561

Description

Given the value of N, you will have to find the value of G. The meaning of G is given in the following code

G=0; 

for(i=1;i&ltN;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 20000 lines of inputs. Each line contains an integer N (1<N <1000001). 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

Source

Contest for 2010 lecture II

题意：给出数字n，求对于所有满足1<= i < j <= n 的所有数对，（i，j）所对应的gcd（i，j）之和。

思路：设f（n） = gcd（1，n）+gcd（2，n）+gcd（3，n）+。。。+gcd（n-1，n）。则所求答案s（n）=f（2）+f（3）+f（4）+。。。+f（n）。

注意到所有的gcd（x，n）的值都是n的约数，所以可以按照这个约数来进行分类。用g（n，i）来表示满足gcd（x，n）=i且x<n的正整数x的个数。则f（n）={i×g（n，i）|i为n的约数}。注意，g（n，i）=phi（n，i）。

如果对于每个n都枚举i，按照数据范围来看肯定会超时，我们不如反着来，先枚举i，再找i的倍数n。这样时间复杂度会进一步减少。

 /*
  * Author:  Joshua
  * Created Time:  2014年09月07日 星期日 19时26分30秒
  * File Name: fzu1969.cpp
  */
 #include<cstdio>
 #include<cstring>
 typedef long long LL;
 #define maxn 1000005
 int phi[maxn],a[maxn],f[maxn];
 LL s[maxn];
 bool p[maxn];
 int tot=;

 void pri()
 {
     memset(p,,sizeof(p));
     for (int i=;i<maxn;++i)
         if (p[i])
         {
             for (int j=i<<;j<maxn;j+=i)
                 p[j]=false;
         }
     for (int i=;i<maxn;++i)
         if (p[i]) a[++tot]=i;
 }

 void phi_table()
 {
     for (int i=;i<maxn;++i)
     {
         phi[i]=i;
         int temp=i;
         for (int j=;j<=tot;++j)
         {
             if (temp==) break;
             if (p[temp])
             {
                 phi[i]/=temp;
                 phi[i]*=temp-;
                 break;
             }
             if (temp % a[j] == )
             {
                 phi[i]/=a[j];
                 phi[i]*=a[j]-;
                 while (temp%a[j]==) temp/=a[j];
             }
         }
     }
 }

 void solve()
 {
     pri();
     phi_table();
     for (int i=;i<maxn;++i)
         for (int j=i;j<maxn;j+=i)
             f[j]+=i*phi[j/i];
     for (int i=;i<maxn;++i)
         s[i]+=s[i-]+f[i];
 }

 int main()
 {
     int n;
     solve();
     while (scanf("%d",&n) && n)
         printf("%lld\n",s[n]);
     return ;
 }