Bzoj 2818 GCD (Euler function)

Title Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=37161

Test instructions: gcd(x, y) = 质数, 1 <= x, y <= n the logarithm of

Idea: Obviously (x, y) = k, 1 <= x, y <= n the logarithm of gcd is equal to (x, y) = 1, 1 <= x, y <= n/k the logarithm of seeking. So, enumerate each prime p number and then ask (x, y) = 1, 1 <= x, y <= n/p for GCD. So the crux of the matter is how to ask gcd (x, y) = 1, 1 <= x, y <= n/pi, [1, y] and the y number of coprime, phi(y) if we make x < y , then the answer is sigma(phi(y)) . Because x, y it is equivalent, so the answer is * *, and because (1, 1) there are only a pair, so-1. The final answer is sigma(sigma(phi(n/prime[i])) * 2 - 1) .

Code

1#include <cstdio>2#include <cstring>3typedefLong LongLL;4 Const intMAXN =10000005;5 6 LL PHI[MAXN];7 intPRIMES[MAXN];8 BOOL  is[MAXN];9 intCNT;Ten  One  A voidInitintN) - { -phi[1] =1L; theCNT =0; -Memset is,false,sizeof( is)); -      for(inti =2; I <= N; ++i) { -         if(! is[i]) { +primes[cnt++] =i; -Phi[i] = i-1; +         } A          for(intj =0; J < cnt && I * primes[j] <= N; ++j) { at              is[i * primes[j]] =true; -             if(i% primes[j] = =0) Phi[i * Primes[j]] = phi[i] *Primes[j]; -             ElsePhi[i * Primes[j]] = phi[i] * (PRIMES[J)-1); -         } -     } - } in  - intMain () to { +     intN; -      while(SCANF ("%d", &n)! =EOF) { the init (n); *LL ans =0; $          for(inti =2; I <= N; ++i) Phi[i] + = phi[i-1]; Panax Notoginseng          for(inti =0; I < CNT; ++i) { -Ans + = phi[n/primes[i]] *2-1; the         } +printf"%lld\n", ans); A     } the     return 0; +}

Bzoj 2818 GCD (Euler function)