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：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 bool is_prime[10000005];
 5 vector<int> prime;
 6 int num[10000005], ans[10000005];
 7 void sieve(int n)
 8 {
 9     for (int i = 0; i <= n; i++) is_prime[i] = true;
10     is_prime[0] = is_prime[1] = false;
11     for (int i = 2; i <= n; i++)
12     {
13         if (is_prime[i])
14         {
15             prime.push_back(i);
16             for (int j = 2 * i; j <= n; j += i) is_prime[j] = false;
17         }
18     }
19 }
20 int main()
21 {
22     sieve(10000000);
23     int n, m, x, a, b;
24     while (scanf("%d", &n) != EOF)
25     {
26         memset(num, 0, sizeof num);
27         memset(ans, 0, sizeof ans);
28         int l = prime.size();
29         for (int i = 0; i < n; i++) { scanf("%d", &x); num[x]++; }
30         for (int i = 0; i < l; i++) // 这段代码和素数筛法是一样的
31         {
32             int now = prime[i];
33             for (int j = now; j <= 10000000; j += now)
34             {
35                 if (num[j]) ans[now] += num[j];
36             }
37         }
38         for (int i = 1; i <= 10000000; i++) ans[i] += ans[i - 1];
39         scanf("%d", &m);
40         for (int i = 0; i < m; i++)
41         {
42             scanf("%d %d", &a, &b);
43             if (a > b) { puts("0"); continue; }
44             a = min(a, 10000000); b = min(b, 10000000);
45             printf("%d\n", ans[b] - ans[a - 1]);
46         }
47     }
48     return 0;
49 }

转载于:https://www.cnblogs.com/wangyiming/p/8909460.html

CF385C Bear and Prime Numbers相关推荐

CF385C Bear and Prime Numbers 数学
题意翻译给你一串数列a.对于一个质数p,定义函数f(p)=a数列中能被p整除的数的个数.给出m组询问l,r,询问[l,r]区间内所有素数p的f(p)之和. 题目描述 Recently, the be ...
codeforces 385C Bear and Prime Numbers
题意:给你一个数列,和m个询问,求数组种 l -r 中所有质数的的倍数的个数和. 解题思路:变形筛法.注意细节解题代码: 1 // File Name: 385c.cpp 2 // Author: ...
Codeforces 385 C Bear and Prime Numbers
题目链接~~> 做题感悟:这题属于想法题,比赛时直接做的 D 题.可是处理坐标处理的头晕眼花的结果到最后也没AC. 解题思路: 由于查询的时候仅仅考虑素数,so~我们仅仅考虑素数就能够,这就须要 ...
Codeforces A - Bear and Prime 100（交互题）
A - Bear and Prime 100 思路:任何一个合数都可以写成2个以上质数的乘积.在2-100中,除了4,9,25,49外都可以写成两个以上不同质数的乘积. 所以打一个质数加这四个数的表: ...
Sum of Consecutive Prime Numbers POJ - 2739（线性欧拉筛+尺取法）
题意: 一些正整数可以由一个或多个连续质数的总和表示.给定一个的正整数n,问满足条件的有多少种情况? 题目: Some positive integers can be represented by ...
7-24 First 50 Prime Numbers (10 分)
7-24 First 50 Prime Numbers Your program reads one natural numbers n in, and prints out the sum of t ...
poj 2739 Sum of Consecutive Prime Numbers
悲剧,如此水题,竟搞了许久.唉,只因题目没看清. 准确的说:应该是准备工作没做好就开始敲代码了!! 下次一定要想的很清楚了再code ing ! //144K 63MS #include <st ...
HOJ 1015 Nearly prime numbers
代码 //Nearly prime number is an integer positive number for which it is possible //to find such prim ...
codeforces 679A Bear and Prime 100 交互
第一次交互题,记录一下吧 #include <cstdio> #include <iostream> #include <ctime> #include <v ...

CF385C Bear and Prime Numbers

CF385C Bear and Prime Numbers相关推荐

最新文章

热门文章