UVA GCD

discription

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

貌似是蓝书上有的一道题，当时刘汝佳是用 N log N 的筛法筛的，但是我们如果把积性函数推出来的话，可以

把那个log也去掉，做到O(N)预处理，O(1)查询。

大概最后就是推这么个积性函数： f(T)=Σφ(d)*(T/d)  ,其中d|T

优化了一个log之后艹爆了时限hhh

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define ll long long
#define maxn 4000005
using namespace std;
int zs[maxn/4],t=0,low[maxn+5];
ll f[maxn+5],n,T;
bool v[maxn+5];

inline void init(){
    f[1]=1,low[1]=1;
    for(int i=2;i<=maxn;i++){
        if(!v[i]) zs[++t]=i,f[i]=i*2-1,low[i]=i;
        for(int j=1,u;j<=t&&(u=zs[j]*i)<=maxn;j++){
            v[u]=1;
            if(!(i%zs[j])){
                low[u]=low[i]*zs[j];
                if(low[i]==i) f[u]=f[i]*zs[j]+low[i]*(zs[j]-1);
                else f[u]=f[i/low[i]]*f[low[u]];
                break;
            }
            
            low[u]=zs[j];
            f[u]=f[i]*(2*zs[j]-1);
        }
    }
    
    for(int i=1;i<=maxn;i++) f[i]+=f[i-1];
}

int main(){
    init();
    while(scanf("%lld",&n)==1&&n) printf("%lld
",f[n]-n*(n+1)/2);    
    return 0;
}