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.
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.
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
2
3
4
Sample Output
1
2
3
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，a,b互质，输出最大的 LCM(a, b），即a，b的最小公倍数最大。
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=;
for(i=; i<=n/; 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 ;
}
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=;
if (n==) printf("1\n");
else
{
if (n%==)
{
max=n/;
if (max%==) max=(max+)*(max-);
else max=(max+)*(max-);
}
else
{
max=n/;
max=max*(max+);
}
printf("%I64d\n",max);
}
}
return ;
}