Whether hdu 3826 number theory n can be filled with a factor is a very good question for the number of integers.

Squarefree number
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 1404 Accepted Submission (s): 373

Problem Description
In mathematics, a squarefree number is one which is divisible by no perfect squares, limit T 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.
 

Input
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
 

Output
For each test case, output the case number first. Then output "Yes" if N is squarefree, "No" otherwise.
 

Sample Input
2
30
75
 

Sample Output
Case 1: Yes
Case 2: No
 

Author
Hanshuai
 
Question:
Can n be rounded out by the number of shards containing a factor?
Can output no, cannot output yes

[Cpp] view plaincopy
/* If a number n can be divisible by the square of a number, the number must be rounded out by the square of a prime number.
If an even number is exceeded by the square of an even number, because the even number = 2 * x must contain the prime number 2, it can be rounded out by the square of quality 2.
If a number can be divisible by an odd number of squares = a prime number x, it can be rounded out by the Quality square.
The data in this question is large and can be processed as follows:
If n> 10 ^ 6
1. Find all the prime numbers in 10 ^ 6 to see if n can be divided into the square of these prime numbers. If No can be output
2. the maximum data size is 10 ^ 18. If there is at most one number x, x is used to integer n because n * n is greater than 10 ^ 18.
Therefore, if it is available to its users, No is output.
 
Extended thinking:
The number of workers can be divided into the square of a certain Prime Number * some prime numbers
*/
# Include <stdio. h>
# Include <math. h>
# Include <string. h>
Int vis [1000000 + 5];
Int prime [79000];
Int cnt [79000 + 5];
Int c;
Void get_prime ()
{
Int I, j, n, m;
C = 0;
N = 1000000;
M = (int) sqrt (n + 0.5 );
Memset (vis, 0, sizeof (vis ));
For (I = 2; I <= m; I ++)
If (! Vis [I])
{
For (j = I * I; j <= n; j + = I)
Vis [j] = 1;
}
For (I = 2; I <= n; I ++) if (! Vis [I])
Prime [c ++] = I;
}
# Include <stdio. h>
Int main ()
{
Int I, j, cas, flag, k;
_ Int64 n, m;
Double num;
Get_prime ();
Scanf ("% d", & cas );
For (k = 1; k <= cas; k ++)
{
Scanf ("% I64d", & n );
Memset (cnt, 0, sizeof (cnt); flag = 0;
For (j = 0; j <c; j ++)
{
M = n;
While (m % prime [j] = 0)
{
M/= prime [j];
Cnt [j] ++;
If (cnt [j]> = 2) {flag = 1; break ;}
}
If (flag) break;
}
If (n> 1000000)
{
Num = sqrt (double) n); // double must be added here; otherwise, the compilation is incorrect.
If (floor (num + 0.5) = num) flag = 1;
} Www.2cto.com
If (flag) printf ("Case % d: No \ n", k );
Else printf ("Case % d: Yes \ n", k );
}
Return 0;
}

 

Author: hnust_xiehonghao