TOJ 3259.Mysterious Number（埃式筛法）

题目链接：http://acm.tju.edu.cn/toj/showp3259.html

3259.   Mysterious Number
Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 1356   Accepted Runs: 435

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
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题意很好理解，所谓神秘数就是指一个数是否能够被它的约数整除。然后我就傻乎乎的枚举暴力来了一波，果不其然——超时。随后学习了一下大神的帖子，知道了一种算法叫：埃式筛法，关于埃式筛法具体的内容见博客：

#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAX = 1000000+2;
int fac[MAX] = {0,1};

int main(){
int low,high,sum;
fill(fac+2,fac+MAX,2);
for (int i = 2; i < (MAX>>1); ++i)
for (int j = 2; i * j < MAX; ++j) {
++fac[i*j];
}
while(~scanf("%d%d",&low,&high)){
sum=0;
for(int i=low;i<=high;i++)
if(i%fac[i]==0)
sum++;
printf("%d\n",sum);
}
}