HDU 4627 (Maximum Least public multiple), hdu4627

HDU 4627 (Maximum Least public multiple), hdu4627
HDU 4627Time Limit:1000 MSMemory Limit:32768KB64bit IO Format:% I64d & % I64u

Description

There are too unsolvable problem in the world. It cocould be about one or about zero. But this time it is about bigger number.
Given an integer n (2 <= n <= 10 9). We shoshould find a pair Positive integerA, B so that a + B = n and [a, B] is as large as possible. [a, B] denote the least common multiplier of a, B.

Input

The first line contains integer T (1 <= T <= 10000), denote the number of the test cases.
For each test cases, the first line contains an integer n.

Output

For each test cases, Print the maximum [a, B]In a line.

Sample Input

3234

Sample Output

123 in this case, because a number n is obtained from two positive integers a + B, you can first determine the range of a and B, it is from 1 to n/2. Because from n/2 + 1 to n, it is the same as the first half. No calculation is required. Then, the division of a and B is used, the maximum output value is LCM (a, B), that is, the least common multiple of a and B is the maximum. Note: This algorithm is easy to understand, but it is easy to time out. When you do it yourself, it is # include <stdio. h>
Int main ()
{
Int a, B, c;
Int max, min, t1, t2;
Int I, n, T;
Scanf ("% d", & T );
While (T --)
{
Scanf ("% d", & n );
Max = 0;
For (I = 1; I <= n/2; I ++)
{
A = I;
B = n-I;
A> B? T1 = a: t1 = B;
T2 = n-t1;
While (t2)
{
C = t1 % t2;
T1 = t2;
T2 = c;
}
Min = a * B/t1;
If (min> max)
Max = min;
}
Printf ("% d \ n", max );
}
Return 0;
}

 

In addition, this is a parity method, which is very simple and clever. It was learned by an ACM expert

# Include <stdio. h>

Int main ()
{
Int n, T;
Long max;
Scanf ("% d", & T );
While (T --)
{
Scanf ("% d", & n );
Max = 0;
If (n = 2) printf ("1 \ n ");
Else
{
If (n % 2 = 0)
{
Max = n/2;
If (max % 2 = 0) max = (max + 1) * (max-1 );
Else max = (max + 2) * (max-2 );
}
Else
{
Max = n/2;
Max = max * (max + 1 );
}
Printf ("% I64d \ n", max );
}
}
Return 0;
}