toj3259[Mysterious Numbers]埃式筛法

最新推荐文章于 2019-07-11 21:19:16 发布

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本文介绍了一种名为“神秘数”的数学概念，这类数能够被其自身不同因数的数量整除。文章提出了一种高效的算法——埃式筛法，用于找出指定范围内所有神秘数的数量，并通过动态规划进行了加速。

3259. Mysterious Number

Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 1267 Accepted Runs: 400

Mysterious Number refers to a number which can be divisible by the number of distinct factors that it has. For instance, 1 (1 factor), 12 (6 factors) and 9 (3 factors) are Mysterious Numbers, but 7(2 factors) or 16 (5 factors) are not.

Given two integers low and high, please calculate the number of Mysterious Numbers between low and high, inclusive.

Input

For each test case, there are two integers low and high in one line separated by spaces.1 ≤ low ≤ high ≤ 1,000,000

Output

Print out the number of Mysterious Numbers between low and high, inclusive.

Sample Input

1 10
10 15

Sample Output

4
1

Author: WTommy

Source: TJU Team Selection Contest 2009 (4)

Submit List Runs Forum Statistics 题目大意是有一种数称为"神秘数", 特征是可以被自己约数的个数整除.比如9有1,3,9三个约数,约数个数为3,而9又能被3整除,所以9是神秘数.又如12有6个约数,12可以被6整除,那12也是神秘数.7有2个约数,但7不能被2整除,所以不是神秘数.
输入给出上下限low, high让你求[low,high]区间有多少个"神秘数".
我靠, 这题搁以前我一定枚举.但是.....呃.....呃..
好吧啊,这题就是埃式筛法.
两个循环可以求出1~100万所有数的约数个数.然后可以dp加速一下.转移
dp[i] = { dp[i-1], i不是神秘数
{ dp[i-1]+1, i是神秘数
然后每次询问low,high.输出dp[high] - dp[low-1]即可.
细节注意:0和1的约数个数处理.

0'00.08"

8672K

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAX = 1000000+1111;
int a[MAX] = {0,1};
int dp[MAX] = {0};
int main() {
	fill(a+2, a+MAX, 2);
	int i, j;
    for (i = 2; i < (MAX>>1); ++i)
        for (j = 2; i * j < MAX; ++j) {
            ++a[i*j];
        }
    //printf("%d\n", a[57896]); //57896 have 8 primes~
	for (i = 1; i < MAX; ++i) {
        dp[i] = dp[i-1] + (i % a[i] ? 0 : 1);
	}
	int low, high;
	while (~scanf(" %d %d", &low, &high)) {
		printf("%d\n", dp[high] - dp[low-1]);
	}
	return 0;
}