CF385C Bear and Prime Numbers

思路：

需要对埃氏筛法的时间复杂度有正确的认识（O(nlog(log(n)))）,我都以为肯定超时了，结果能过。

实现：

#include <bits/stdc++.h>
using namespace std;

bool is_prime[10000005];
vector<int> prime;
int num[10000005], ans[10000005];
void sieve(int n)
{
    for (int i = 0; i <= n; i++) is_prime[i] = true;
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i <= n; i++)
    {
        if (is_prime[i])
        {
            prime.push_back(i);
            for (int j = 2 * i; j <= n; j += i) is_prime[j] = false;
        }
    }
}
int main()
{
    sieve(10000000);
    int n, m, x, a, b;
    while (scanf("%d", &n) != EOF)
    {
        memset(num, 0, sizeof num);
        memset(ans, 0, sizeof ans);
        int l = prime.size();
        for (int i = 0; i < n; i++) { scanf("%d", &x); num[x]++; }
        for (int i = 0; i < l; i++) // 这段代码和素数筛法是一样的
        {
            int now = prime[i];
            for (int j = now; j <= 10000000; j += now)
            {
                if (num[j]) ans[now] += num[j];
            }
        }
        for (int i = 1; i <= 10000000; i++) ans[i] += ans[i - 1];
        scanf("%d", &m);
        for (int i = 0; i < m; i++)
        {
            scanf("%d %d", &a, &b);
            if (a > b) { puts("0"); continue; }
            a = min(a, 10000000); b = min(b, 10000000);
            printf("%d
", ans[b] - ans[a - 1]);
        }
    }
    return 0;
}