Light OJ 1116 Ekka Dokka（数学，二进制的应用）

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 of M 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, where W 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). And W 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 print Nand M. If there are multiple solutions, then print the result where M 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

 

思路：

例如样例中的10，转换为二进制则是1010，要找一对N（奇数）*M（偶数），并要后者尽量小，则后者应该就要尽量大，经过草稿纸上的研究发现（证明如下），把二进制中的末尾的0全部去掉就得到了一个最大的奇数，即把1010变为101，则转换为十进制是则是5，那么另一个数偶数就是2，第二第三个样例也如此。

接上证明：

假设有一个整数的二进制表示是：AB

那么转换为十进制应该是A*2^1+B*2^0，能够整除后面B*2^0，那么算出一个最大的奇数，就能相应地得出一个偶数；

（以上是本人的一些弱渣证明思路，如有错误，请大胆指出，一定虚心接受）

AC代码：

#include
#include
#include
using namespace std;
long long two[100];
int main()
{
  two[0]=1;two[1]=2;
  for(int i=2;i<=64;i++)
  {
    two[i]=two[i-1]*2;
  }
  int t,ca=1;
  cin>>t;
  while(t--)
  {
    long long w,n,m;
    cin>>w;
    long long x=w,a[100],i=0;
    while(x)
    {
      i++;
      a[i]=x%2;
      x/=2;
    }
    long long k;
    for(k=1;k<=i;k++)
    {
      if(a[k])break;
    }
    if(k==1&&a[k])
    {
      cout<<"Case "<=k;j--)
    {
      if(a[j])ans+=two[j-k];
    }
    cout<<"Case "<