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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：

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 typedef long long LL;
 7 
 8 const int maxn = 1000000+3;
 9 LL G[maxn];
10 int opl[maxn];
11 void init()
12 {
13     LL i,j;
14     for(i=2;i<maxn;i++) opl[i] = i;
15     for(i=2;i<maxn;i++)
16     {
17         if(opl[i]==i)
18         {
19             for(j=i;j<maxn;j=j+i)
20                 opl[j]=opl[j]/i*(i-1);
21         }
22         for(j=1;i*j<maxn;j++)
23             G[j*i] = G[j*i] + opl[i]*j;
24     }
25     for(i=3;i<maxn;i++)
26         G[i] +=G[i-1];
27 }
28 int main()
29 {
30     init();
31     int T,n;
32     while(scanf("%d",&n)>0)
33     {
34         printf("%lld\n",G[n]);
35     }
36     return 0;
37 }

View Code

AC代码: 

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 typedef long long LL;
 7 
 8 const int maxn = 1e6+3;
 9 int phi[maxn];
10 LL g[maxn];
11 void init()
12 {
13     for(int i=1;i<maxn;i++) phi[i] = i;
14     for(int i=2;i<maxn;i++)
15     {
16         if(phi[i]==i) phi[i] = i-1;
17         else continue;
18         for(int j=i+i;j<maxn;j=j+i)
19             phi[j] = phi[j]/i*(i-1);
20     }
21     for(int i=1;i<maxn;i++) g[i] = phi[i];
22     for(int i=2;i<=1000;i++)
23     {
24         for(LL j=i*i,k=i;j<maxn;j=j+i,k++)
25         if(i!=k)
26             g[j] = g[j] + phi[i]*k + phi[k]*i;
27         else g[j] = g[j] + phi[i]*k;
28     }
29     g[1] = 0;
30     for(int i=2;i<maxn;i++) g[i] = g[i]+g[i-1];
31 }
32 int main()
33 {
34     init();
35     int T,n;
36     scanf("%d",&T);
37     while(T--)
38     {
39         scanf("%d",&n);
40         printf("%lld\n",g[n]);
41     }
42     return 0;
43 }

spoj 3871. GCD Extreme 欧拉+积性函数

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原文地址：http://www.cnblogs.com/tom987690183/p/3946212.html