BZOJ 2818 GCD（欧拉函数）

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=37161

题意：gcd(x, y) = 质数, 1 <= x, y <= n的对数

思路：显然gcd(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gcd(x, y) = 1, 1 <= x, y <= n/p的个数。那么问题的关键就是怎么求gcd(x, y) = 1, 1 <= x, y <= n / pi，在[1, y]和y互质的数有phi(y)个，如果我们令x < y，那么答案就是sigma(phi(y))。因为x, y是等价的，所以答案*2，又因为(1, 1)只有一对，所以-1。最终答案为sigma(sigma(phi(n/prime[i])) * 2 - 1)。

code：

 #include <cstdio>
 #include <cstring>
 typedef long long LL;
 const int MAXN = ;

 LL phi[MAXN];
 int primes[MAXN];
 bool is[MAXN];
 int cnt;

 void init(int n)
 {
     phi[] = 1L;
     cnt = ;
     memset(is, false, sizeof(is));
     for (int i = ; i <= n; ++i) {
         if (!is[i]) {
             primes[cnt++] = i;
             phi[i] = i - ;
         }
         for (int j = ; j < cnt && i * primes[j] <= n; ++j) {
             is[i * primes[j]] = true;
             if (i % primes[j] == ) phi[i * primes[j]] = phi[i] * primes[j];
             else phi[i * primes[j]] = phi[i] * (primes[j] - );
         }
     }
 }

 int main()
 {
     int n;
     while (scanf("%d", &n) != EOF) {
         init(n);
         LL ans = ;
         for (int i = ; i <= n; ++i) phi[i] += phi[i - ];
         for (int i = ; i < cnt; ++i) {
             ans += phi[n / primes[i]] *  - ;
         }
         printf("%lld\n", ans);
     }
     return ;
 }