HDU 3826 Squarefree number 判断无平方因子的数

Squarefree number

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1636    Accepted Submission(s): 438


[align=left]Problem Description[/align]
In mathematics, a squarefree number is one which is divisible by no perfect squares, except 1. For example, 10 is square-free but 18 is not, as it is divisible by 9 = 3^2. Now you need to determine whether an integer is squarefree or not.
 
 

[align=left]Input[/align]
The first line contains an integer T indicating the number of test cases.
For each test case, there is a single line contains an integer N.

Technical Specification

1. 1 <= T <= 20
2. 2 <= N <= 10^18
 
 

[align=left]Output[/align]
For each test case, output the case number first. Then output "Yes" if N is squarefree, "No" otherwise.
 
 

[align=left]Sample Input[/align]

2 30 75

 
 

[align=left]Sample Output[/align]

Case 1: Yes Case 2: No
 
思路：题意说的是，给你一个数N，问这个数是否存在因子且该因子是某个整数的平方，若存在则输出No,否则输出Yes.由于N的最大范围达到10^18，这里可以用筛选法得到2-10^6之间的素数，然后依次用小于N的这些数去除N，若某个素数可以除N两次以上，则输出No；否则，若除完2-10^6之间的素数后，N为1则输出Yes；最后若N不为1，则将N开根，得到的数若是整数，输出No，否则输出Yes。简单的原理就是，如果N不存在10^6一下的素数因子，则N必定是大于10^6的数，它可能是某个大于10^6的素数，也可能是大于10^6的某个数的平方（不可能是三次方，也不存在平方后再乘以小于10^6的数，因为之前已经筛选掉了。)
 




View Code

1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4 #include <cstring>
5 #include <algorithm>
6
7 using namespace std;
8
9 int prime[1000001];
10 bool vis[1000001];
11
12 int main()
13 {
14     memset(vis,0,sizeof(vis));
15     int idx = 0;
16     for(int i=2; i<=1000; i++)
17     {
18         if(!vis[i])
19         {
20             for(int j=i*i; j<=1000000; j+=i)
21                 vis[j] = true;
22         }
23
24     }
25     for(int i=2; i<=1000000; i++)
26         if(!vis[i])
27             prime[idx++] = i;
28     int Case;
29     scanf("%d",&Case);
30     for(int ca = 1; ca <= Case; ca ++)
31     {
32         __int64 n;
33         scanf("%I64d",&n);
34         printf("Case %d: ",ca);
35         int time = 0;
36         for(int i=0; i<idx; i++)
37         {
38             time = 0;
39             while(n % prime[i] == 0)
40             {
41                 n /= prime[i];
42                 time ++;
43             }
44             if(time >= 2)
45             {
46                 break;
47             }
48         }
49         if(time >= 2)
50         {
51             printf("No\n");
52             continue;
53         }
54         if(n == 1)
55         {
56             printf("Yes\n");
57             continue;
58         }
59         if(((__int64)sqrt(n*1.0+0.00000001)*(__int64)sqrt(n*1.0+0.00000001)) == n)
60             printf("No\n");
61         else
62             printf("Yes\n");
63     }
64     return 0;
65 }