HDU 4627

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Description

There are many unsolvable problem in the world.It could be about one or about zero.But this time it is about bigger number. 
Given an integer n(2 <= n <= 10 ⁹).We should find a pair of positive integer a, 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

3
2
3
4

Sample Output

1
2
3

此题的话，因为一个数n由两个正整数a+b得来，所以可以先确定a和b的范围，是从1到n/2

因为从n/2+1到n，是和前半部分重复，不用计算

然后，就用辗转相除法，算出a和b的最大公约数，a*b/最大公约数=最小公倍数 。

注意：这种算法易懂，但是很容易超时，自己做的时候就是

#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;
}

另外还有一种，这是一种奇偶求法，很简单巧妙，经某ACM大神指点得知

#include <stdio.h>

int main()
{
int n,T;
long 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;
}