Logarithm of greatest common divisor as prime number in Bzoj 2818:gcd interval (Application of Euler function)

Portal
2818:gcd

Time Limit:10 Sec Memory limit:256 MB
submit:3649 solved:1605
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Description

Given an integer n, 1<=x,y<=n and gcd (x, y) are prime
Number pairs (x, y) how many pairs.

Input

An integer n

Output

Title

Sample Input

4

Sample Output

4
HINT

Hint

For examples (2,2), (2,4), (3,3), (4,2)

1<=n<=10^7

Source

Hubei Province Team Mutual test

Problem Solving Ideas:
The problem is that the number of <=n gcd (x, y) = = primes (2,4) and (4,2) are considered to be different, so we can think of enumerating each prime number so that it gcd (x, y) =p, then we can think of the count of the 1,y/p in [y/p] (where the default Y >X), then we are asking for a Euler function value, then we extend it to the 1-n interval, that is, the Euler function value [1,n/p], but we need to ask for the SIGMAEUALR (n/p) prefix and because Y is taken from the 1-n, so the logarithm is sum[n/p ]*2-1, because it is logarithmic, and there are repeated cases (the case itself is a prime number)

See the code for details:

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>usingNamespace Std;typedefLong LongLL;ConstLL MAXN =1e7+5;BOOLPRIME[MAXN];///   tag array is not a prime numberLL PHI[MAXN];///   Euler function values, I Euler value =phi[i]LL P[MAXN];  The value of the element factorLL cnt =0;voidGet_phi ()///   sieve method to find Euler function{cnt =0; memset (Prime,true,sizeof(prime)); phi[1] =1; for(LL i=2; i<maxn; i++)///   Linear sieve method{if(Prime[i])///   Prime{p[cnt++] = i; Phi[i] = i1;  The Euler function value of the prime number is prime-1} for(LL j=0; j<cnt; J + +) {if(I*p[j] > MAXN) Break; PRIME[I*P[J]] =false;///   multiples of prime number, so i*p[j] is not a prime number            if(I%p[j] = =0)/   //nature: I mod p = = 0, then phi (i * p) = = p * PHI (i){Phi[i*p[j]] = p[j] * Phi[i]; Break; }ElsePHI[I*P[J]] = (p[j]-1) * Phi[i];///  i mod p! = 0, then phi (i * p) = = Phi (i) * (p-1)}}}ll SUM[MAXN];  ///prefix andvoidGet_sum () {memset (sum,0,sizeof(sum)); for(LL i=1; i<maxn; i++) Sum[i] = sum[i-1]+phi[i];}intMain () {Get_phi ();    Get_sum (); LL N; while(~SCANF ("%lld", &n)) {LL ans =0; for(LL i=0; i<cnt&&p[i]<=n; i++) {ans = ans+sum[n/p[i]]*2-1; } printf ("%lld\n", ans); }return 0;}

Logarithm of greatest common divisor as prime number in Bzoj 2818:gcd interval (Application of Euler function)