UVa 11426

首先我们了解一下欧拉函数phi[x],phi[x]表示的是不超过x且和x互素的整数个数

phi[x]=n*(1-1/p1)*(1-1/p2)....(1-1/pn);

计算欧拉函数的代码为

 int euler_phi(int n){
     int m=(int)sqrt(n+0.5);
     int ans=n;
     for(int i=;i*i<=m;i++){
         if(n%i==){//
             ans=ans/i*(i-);
             while(n%i==)n=n/;
         }
     }
     if(n>)ans=ans/n*(n-);
     return ans;
 }

也可以通过类似于素数打表的方法来对欧拉函数进行打表

 void phi_table(int n){
     for(int i=;i<=n;i++)phi[i]=;
     phi[]=;
     for(int i=;i<=n;i++){
         if(phi[i]==)
         for(int j=i;j<=n;j=j+i){
             if(phi[j]==)phi[j]=j;
             phi[j]=phi[j]/i*(i-);
         }
     }
 }

题意

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

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

设f(n)=gcd(1,2)+gcd(1,3)+gcd(2,3)+....+gcd(n-1,n); 题意是求1<=i<j<=n的数对（i,j）所对应的gcd(i,j)的和s(j)

s(n)=f(1)+f(2)+f(3)+f(4)+....f(n);

当i为n的约数的时候

我们选择t(i)表示gcd(x,n)=i(x<n)的正整数的个数 则f(n)=sum{i*t(i)}

gcd(x,n)=i<==>gcd(x/i,n/i)=1;

满足下t(i)为phi[n/i];(感觉还是可以理解的）

在代码实现中，如果枚举（1-n）的约数的话明显效率慢，所以我们可以枚举i的倍数来更新f(n);

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<map>
 #include<set>
 #include<vector>
 #include<cstdlib>
 #include<string>
 #define eps 0.000000001
 typedef long long ll;
 typedef unsigned long long LL;
 using namespace std;
 const int N=+;
 ll S[N],f[N];
 ll phi[N];
 ll euler_phi(int n){
     int m=(int)sqrt(n+0.5);
     ll ans=n;
     for(int i=;i*i<=m;i++){
         if(n%i==){//2一定是素因子
             ans=ans/i*(i-);
             while(n%i==)n=n/;
         }
     }
     if(n>)ans=ans/n*(n-);
     return ans;
 }
 void phi_table(int n){
     for(int i=;i<=n;i++)phi[i]=;
     phi[]=;
     for(int i=;i<=n;i++){
         if(phi[i]==)
         for(int j=i;j<=n;j=j+i){
             if(phi[j]==)phi[j]=j;
             phi[j]=phi[j]/i*(i-);
         }
     }
 }
 void init(){
     int n=N;
     phi_table(N);
     memset(f,,sizeof(f));
     for(int i=;i<=n;i++){
         for(int j=i*;j<=n;j=j+i)f[j]=f[j]+i*phi[j/i];
     }
     memset(S,,sizeof(S));
     S[]=f[];
     for(int i=;i<=n;i++)S[i]=S[i-]+f[i];
 }
 int main(){
     int n;
     init();
     while(scanf("%d",&n)!=EOF){
         //init(n);
         if(n==)break;
         printf("%lld\n",S[n]);
     }
 }