bzoj 2818 Gcd 【欧拉函数】

问题：求gcd(x,y)==质数, 1<=x,y<=n的有多少对？

做这题的时候，懂得了一个非常重要的转化：求(x, y) = k, 1 <= x, y <= n的对数等于求(x, y) = 1, 1 <= x, y <= n/k的对数！所以，枚举每个质数p，然后求(x, y) = 1, 1 <= x, y <= n/p的个数。

(x, y) = 1 的个数如何求呢？欧拉函数！

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#include <time.h>
#include <queue>
#include <iterator>

using namespace std;

const int MAXN = 1000000;
int n;

int main()
{
  while (scanf("%d", &n) != EOF)
  {
    bool com[MAXN];
    int primes = 0, prime[MAXN], phi[MAXN];

    phi[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
      if (!com[i])
      {
        prime[primes++] = i;
        phi[i] = i - 1;
      }
      for (int j = 0; j < primes && i*prime[j] <= n; ++j)
      {
        com[i*prime[j]] = true;
        if (i % prime[j])
          phi[i*prime[j]] = phi[i] * (prime[j] - 1);
        else
        {
          phi[i*prime[j]] = phi[i] * prime[j]; break;
        }
      }
    }

    for (int i = 2; i <= n; i++)
      phi[i] = phi[i] + phi[i-1];
    long long ans = 0;

    for (int i = 0; i < primes; i++)
      ans += phi[n/prime[i]] * 2 -1;
    printf("%lld\n",ans);
  }
  return 0;
}