Spoj 3871. GCD extreme Euler + product function

3871. GCD extremeproblem 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. = ""

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

Question:

G = 0;

For (k = I; k <n; k ++)

For (j = I + 1; j <= N; j ++)

{

G + = gcd (K, J );

}

Idea: G [N] = sigma (d | n Phi [d] * (N/D); this can be used to obtain the value of S [N], and the sum can be accumulated.

The key is that the G [N] function can be filtered because it is a product function.

Two filtering methods: TLE and AC.

Timeout code:

 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 code:

 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 Euler + product function