UVA 11426 GCD - Extreme (II) （欧拉函数）

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Problem J
GCD Extreme (II)
Input: Standard Input

Output: Standard Output

 

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  　　 Output for Sample Input

10 100 200000 0

67 13015 143295493160

---

Problemsetter: Shahriar Manzoor

Special Thanks: SyedMonowarHossain

设dp[i]=gcd(1,i)+gcd(2,i)+……+gcd(i-1,i);

则ans[n]=dp[2]+dp[3]+……+dp[n].

由此问题已经转化成如何求dp[i]了，即需要求1到i-1所有数与i的gcd的和。

设k为满足gcd(x,i)=j且x<i的正整数的个数，则dp[i]=∑j*k;

同时，由于gcd(x,i)=j等价于gcd(x/j,i/j)=1,也就是phi[i/j];

接下来反过来求，那就不需要分解素因子了

#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn=4000010;
int phi[maxn];
ll dp[maxn+1];
ll ans[maxn+1];
void phi_table()
{
    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);
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    phi_table();
    for(int i=1;i<maxn;i++)
    {
        for(int j=i*2;j<maxn;j+=i)dp[j]+=(long long)i*(long long)phi[j/i];
    }
    ans[2]=dp[2];
    for(int i=3;i<maxn;i++)ans[i]=ans[i-1]+dp[i];
    int n;
    while(cin>>n&&n)
    {
        cout<<ans[n]<<endl;
    }
    return 0;
}