3871. GCD Extreme

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:101002000000Output:6713015143295493160

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