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1109 - False Ordering


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Time Limit: 1 second(s)

Memory Limit: 32 MB


We define b is a Divisor of a number a if a is divisible by b. So, the divisors of 12 are 1, 2, 3, 4, 6, 12. So, 12 has 6 divisors.

Now you have to order all the integers from 1 to 1000. x will come before y if

1)                  number of divisors of x is less than number of divisors of y

2)                  number of divisors of x is equal to number of divisors of y and x > y.

Input

Input starts with an integer T (≤ 1005), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 1000).

Output

For each case, print the case number and the nth number after ordering.

Sample Input

Output for Sample Input

5

1

2

3

4

1000

Case 1: 1

Case 2: 997

Case 3: 991

Case 4: 983

Case 5: 840





题解：刚开始想到结构体，然后不敢写，然后想了其他办法，果断没想到，就回来做了，真ZZ

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int n;
struct node
{
  int id,shu;
}a[1000010];
bool cmp(node x,node y)
{
  if(x.id!=y.id)
    return x.id<y.id;
  else
    return x.shu>y.shu;
}
void get()
{
  int num=1;
  memset(a,0,sizeof(a));
  for(int i=1000;i>=1;i--)
  {
    int cnt=0;
    for(int j=1;j*j<=i;j++)
    {
      if(i%j==0)
      {
        cnt=cnt+2;
      }
      if(j*j==i)  cnt--;
    }
    a[num].shu=i;
    a[num++].id=cnt;
  }
  sort(a+1,a+1001,cmp);
//  for(int i=1;i<=100;i++)
//    printf("--%d--%d--\n",a[i].id,a[i].shu);
}
int main()
{
  int t,text=0;
  scanf("%d",&t);
  get();
  while(t--)
  {
    scanf("%d",&n);
    printf("Case %d: %d\n",++text,a[n].shu);
  }
  return 0;
}