Uva 11426 - GCD - Extreme (II) 欧拉函数

链接：戳这里

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

            G =       ∑     ∑        GCD(i, j)

i=1    j=i+1

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

题意：输入正整数n,求gcd(1,2)+gcd(1,3)+gcd(2,3)+...+gcd(n-1,n),即对所有的满足(1<=i<j<=n)的数对(i,j)所对应的gcd(i,j)之和。比如当n==4时     ans=gcd(1,2)+gcd(1,3)+gcd(2,3)+gcd(1,4)+gcd(2,4)+gcd(3,4)

先设 f[n]=gcd(1,n)+gcd(2,n)+...+gcd(n-1,n)

那么 ans=f[2]+f[3]+...+f[n]     递推一下sum[n]=sum[n-1]+f[n]

所以现在只需要求f[n]，我们先设g(x,n)=i  表示gcd(x,n)==i的有多少个  显然x肯定是n的约数

其实出现gcd马上就要想到欧拉函数了    gcd(x,n)==i   -> gcd(x/i,n/i)==1  也就是找出phi[n/i]的个数了

但是直接暴力找每次的f[n]的话,肯定是会超时的,所以模拟打表的过程处理f[n]  这里也是技巧啊  涨姿势

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 4000000
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
/*
    令f[n]={   gcd(1,n)+gcd(2,n)+gcd(3,n)+...+gcd(n-1,n)   }
    s[n]=f[2]+f[3]+f[4]....f[n]   s[n]表示答案
    -> s[n]=s[n-1]+f[n];

*/

ll f[4000100],phi[4000100],sum[4000100];
void init(){
    mst(phi,0);
    phi[1]=1;
    for(int i=2;i<=MAX;i++){
        if(phi[i]) continue;
        for(int j=i;j<=MAX;j+=i){
            if(!phi[j]) phi[j]=j;
            phi[j]=phi[j]/i*(i-1);
        }
    }
    mst(f,0);
    for(int i=1;i<=MAX;i++){
        for(int j=i*2;j<=MAX;j+=i){
            f[j]+=(ll)phi[j/i]*i;
        }
    }
    sum[2]=f[2];
    for(int i=3;i<=MAX;i++) sum[i]=sum[i-1]+f[i];
}
int main(){
    init();
    int n;
    while(scanf("%d",&n)!=EOF){
        if(n==0) break;
        printf("%lld\n",sum[n]);
    }
    return 0;
}