UVA-11426-GCD - Extreme (II)-数论-欧拉函数

UVA-11426-GCD - Extreme (II)-数论-欧拉函数

【Description】

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

$$> G=\sum_{i=1}^{n-1}\sum_{j=i+1}^NGCD(i,j) >$$

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 findsthe 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.

---

【Examples】

Sample Input

10
100
200000
0

Sample Output

67
13015
143295493160

---

【Problem Description】

题意就一个式子。不用多说。

【Solution】

数论。欧拉函数。
对于题目给的式子不太好求，我们将其转换一下。转换为以下形式：

$$> G=\sum_{i=2}^{N}\sum_{j=1}^{i-1}GCD(j,i) >$$

可以将两个式子分解开来对比，两个是等价的。
G=sum[2]+sum[3]+...+sum[N].
sum[i]=gcd(1,i)+gcd(2,i)+...+gcd(i-1,i).
我们设gcd(j,i)=x,我们假设所有满足条件的j中(1<=j<i),gcd(j,i)=x的个数有y个(1<=x<=i)。
则sum[i]=x1*y1+x2*y2+...+xn*yn.
对于每个x，x*y=x*Phi[i/x],

---

【Code】

/*
 * @Author: Simon 
 * @Date: 2018-09-03 15:27:18 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-09-03 21:00:51
 */
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 4000005
int prime[maxn];
int Ph[maxn];
int sum[maxn];
void getprime()//欧拉筛
{
    for(int i=2;i<=maxn;i++)
    {
        if(!prime[i])
        {
            prime[++prime[0]]=i;
            Ph[i]=i-1;
        }
        for(int j=1;j<=prime[0]&&i*prime[j]<=maxn;j++)
        {
            prime[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                Ph[i*prime[j]]=Ph[i]*prime[j];
                break;
            }
            else Ph[i*prime[j]]=(prime[j]-1)*Ph[i];
        }
    }
}
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    getprime();
    for(int i=1;i<=maxn;i++)
    {
        for(int j=i+i;j<=maxn;j+=i)
        {
            sum[j]+=i*Ph[j/i];//求sum[i]
        }
    }
    for(int i=1;i<=maxn;i++)
    {
        sum[i]+=sum[i-1];
    }
    int n;
    while(cin>>n&&n)
    {
        cout<<sum[n]<<endl;
    }
    cin.get(),cin.get();
    return 0;
}