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：

 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 }