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[];
 vector<int> prime;
 int num[], ans[];
 void sieve(int n)
 {
     for (int i = ; i <= n; i++) is_prime[i] = true;
     is_prime[] = is_prime[] = false;
     for (int i = ; i <= n; i++)
     {
         if (is_prime[i])
         {
             prime.push_back(i);
             for (int j =  * i; j <= n; j += i) is_prime[j] = false;
         }
     }
 }
 int main()
 {
     sieve();
     int n, m, x, a, b;
     while (scanf("%d", &n) != EOF)
     {
         memset(num, , sizeof num);
         memset(ans, , sizeof ans);
         int l = prime.size();
         for (int i = ; i < n; i++) { scanf("%d", &x); num[x]++; }
         for (int i = ; 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 <= ; j += now)
             {
                 if (num[j]) ans[now] += num[j];
             }
         }
         for (int i = ; i <= ; i++) ans[i] += ans[i - ];
         scanf("%d", &m);
         for (int i = ; i < m; i++)
         {
             scanf("%d %d", &a, &b);
             if (a > b) { puts(""); continue; }
             a = min(a, ); b = min(b, );
             printf("%d\n", ans[b] - ans[a - ]);
         }
     }
     return ;
 }