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 = 10000005;

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


void init(int n)
{
    phi[1] = 1L;
    cnt = 0;
    memset(is, false, sizeof(is));
    for (int i = 2; i <= n; ++i) {
        if (!is[i]) {
            primes[cnt++] = i;
            phi[i] = i - 1;
        }
        for (int j = 0; j < cnt && i * primes[j] <= n; ++j) {
            is[i * primes[j]] = true;
            if (i % primes[j] == 0) phi[i * primes[j]] = phi[i] * primes[j];
            else phi[i * primes[j]] = phi[i] * (primes[j] - 1);
        }
    }
}

int main()
{
    int n;
    while (scanf("%d", &n) != EOF) {
        init(n);
        LL ans = 0;
        for (int i = 2; i <= n; ++i) phi[i] += phi[i - 1]; 
        for (int i = 0; i < cnt; ++i) {
            ans += phi[n / primes[i]] * 2 - 1;
        }
        printf("%lld\n", ans);
    }
    return 0;
}