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

AC代码:

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

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