spoj 3871. GCD Extreme 欧拉+积性函数

3871. GCD Extreme

Problem code: GCDEX

Given the value of N, you will have to find the value of G. The meaning of G is given in the following code

G=0;

for(k=i;k< N;k++)

for(j=i+1;j<=N;j++)

{

G+=gcd(k,j);

}

/*Here gcd() is a function that finds the greatest common divisor of the two input numbers*/

Input

The input file contains at most 20000 lines of inputs. Each line contains an integer N (1<n<1000001). the="" meaning="" of="" n="" is="" given="" in="" problem="" statement.="" input="" terminated="" by="" a="" line="" containing="" single="" zero.="" <h3="">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.

Example

Input:
10
100
200000
0

Output:
67
13015
143295493160

题意：

G=0;

for(k=i;k< N;k++)

for(j=i+1;j<=N;j++)

{

G+=gcd(k,j);

}

思路: G[n] = sigma( d|n  phi[d]*(n/d) ); 这个能求出S[n]的值,累加求和就行。

　　　关键在于G[n]函数能用筛选来做，因为是积性函数。

两种筛选方法，一种TLE，一种ac。

超时代码：

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;

const int maxn = 1000000+3;
LL G[maxn];
int opl[maxn];
void init()
{
    LL i,j;
    for(i=2;i<maxn;i++) opl[i] = i;
    for(i=2;i<maxn;i++)
    {
        if(opl[i]==i)
        {
            for(j=i;j<maxn;j=j+i)
                opl[j]=opl[j]/i*(i-1);
        }
        for(j=1;i*j<maxn;j++)
            G[j*i] = G[j*i] + opl[i]*j;
    }
    for(i=3;i<maxn;i++)
        G[i] +=G[i-1];
}
int main()
{
    init();
    int T,n;
    while(scanf("%d",&n)>0)
    {
        printf("%lld
",G[n]);
    }
    return 0;
}

AC代码: 

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;

const int maxn = 1e6+3;
int phi[maxn];
LL g[maxn];
void init()
{
    for(int i=1;i<maxn;i++) phi[i] = i;
    for(int i=2;i<maxn;i++)
    {
        if(phi[i]==i) phi[i] = i-1;
        else continue;
        for(int j=i+i;j<maxn;j=j+i)
            phi[j] = phi[j]/i*(i-1);
    }
    for(int i=1;i<maxn;i++) g[i] = phi[i];
    for(int i=2;i<=1000;i++)
    {
        for(LL j=i*i,k=i;j<maxn;j=j+i,k++)
        if(i!=k)
            g[j] = g[j] + phi[i]*k + phi[k]*i;
        else g[j] = g[j] + phi[i]*k;
    }
    g[1] = 0;
    for(int i=2;i<maxn;i++) g[i] = g[i]+g[i-1];
}
int main()
{
    init();
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        printf("%lld
",g[n]);
    }
    return 0;
}