uva11426 GCD Extreme (II)

Test instructions: Sum (gcd (i,j), 1<=i<j<=n)1<n<4000001

Ideas:

1. Establish a recursive relationship, S (n) =s (n-1) +gcd (1,n) +gcd (2,n) +......+GCD (n-1,n);

2. Set f (n) =gcd (1,n) +gcd (2,n) +......+GCD (n-1,n).

GCD (X,n) =i is an approximate (x<n) of N, classified according to this approximate. A constraint that satisfies the GCD (x,n) =i has g (n,i), then f (n) =sum (I*g (n,i)).

The GCD (x,n) =i is equivalent to gcd (x/i,n/i) = 1, so g (n,i) is equivalent to Phi (n/i). Phi (x) is an Euler function.

3. Reduce the complexity of time. Pretreatment of phi[x] table by sieve method

By using the Sieve method, the enumeration factor of F (x) is pretreated, and all multiples of the solution are updated.

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cctype>5#include <cstring>6#include <vector>7#include <cassert>8 9 using namespacestd;Ten Const intMAXN =4000010; One #defineLL Long Long A #defineCLC (A, B) memset (A,b,sizeof (a)) - LL S[MAXN],F[MAXN]; - LL PHI[MAXN]; the voidPhi_table (intN) - { -      for(intI=2; i<=n;i++) -phi[i]=0; +phi[1]=1; -      for(intI=2; i<=n;i++) +     { A         if(phi[i]==0) at         { -              for(intj=i;j<=n;j+=i) -             { -                 if(!Phi[j]) -phi[j]=J; -phi[j]=phi[j]/i* (I-1); in             } -         } to     } + } -  the intMain () * { $ phi_table (MAXN);Panax NotoginsengCLC (F,0); -      for(intI=1; i<=maxn;i++) the     { +          for(intn=i*2; n<=maxn;n+=i) Af[n]+=i*phi[n/i]; the     } +s[2]=f[2]; -      for(intn=3; n<=maxn;n++) $s[n]=s[n-1]+F[n]; $     intN; -      while(SCANF ("%d",&N), N) -     { theprintf"%lld\n", S[n]); -     }Wuyi     return 0; the}

View Code

uva11426 GCD Extreme (II)