LightOJ 1116 - Ekka Dokka (简单数学)


1116 - Ekka Dokka

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Ekka and his friend Dokka decided to buy a cake. They both love cakes and that's why they want to share the cake after buying it. As the name suggested that Ekka is very fond of odd numbers and Dokka is very fond of even numbers, they want
to divide the cake such that Ekka gets a share of N square centimeters and Dokka gets a share ofM square centimeters where
N is odd and M is even. Both N and
M are positive integers.
They want to divide the cake such that N * M = W, whereW is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case contains an integer W (2 ≤ W < 263). AndW will not be a power of
2.

Output

For each case, print the case number first. After that print"Impossible" if they can't buy their desired cake. If they can buy such a cake, you have to printN and
M. If there are multiple solutions, then print the result whereM is as small as possible.

Sample Input	Output for Sample Input
3 10 5 12	Case 1: 5 2 Case 2: Impossible Case 3: 3 4

 
题意：给出一个W,要求N*M=W,且N是奇数，M是偶数。求出N和M的值，要求M的值尽可能少。
题解：一个偶数乘以一个奇数的结果必定是偶数，所以当W为奇数时，没有符合条件的N与M的值，输出Impossible。  当W是偶数时，1*W=W。所以必定存在可行解。

代码如下：

#include<cstdio>
#include<cstring>
int main()
{
int t,k=1;
long long w,n,m;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&w);
if(w&1)
printf("Case %d: Impossible\n",k++);
else
{
for(m=2;m<=w;++m)
{
if(w%m==0&&((w/m)&1))
{
n=w/m;
break;
}
}
printf("Case %d: %lld %lld\n",k++,n,m);
}
}
return 0;
}