1109 - False Ordering

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 n^th number after ordering.

题目大意：

　　题目说的是，给你[1,1000]里面的数字，让你按照如下的两条规则来排序，

　　1.如果x的因子的个数大于y的因子的个数，那么y排在x的前面。

　　2.如果x的因子的个数等于y的因子的个数，且x>y，那么x排在y的前面。

解题思路：

　　直接打表，然后按照以上两条规则来sort就可以了。

代码:

 1 # include<cstdio>
 2 # include<iostream>
 3 # include<algorithm>
 4
 5 using namespace std;
 6
 7 # define MAX 1234
 8
 9 int a[MAX];
10
11 struct node
12 {
13     int val;
14     int id;
15 }num[MAX];
16
17
18 int cmp ( const struct node & x,const struct node & y )
19 {
20     if ( x.val==y.val )
21     {
22         return x.id > y.id;
23     }
24     return x.val < y.val;
25 }
26
27
28 void init()
29 {
30     for ( int i = 1;i <= 1000;i++ )
31     {
32         num[i].id = i;
33         for ( int j = 1;j <= i;j++ )
34         {
35             if ( i%j == 0 )
36             {
37                 num[i].val++;
38             }
39         }
40     }
41 }
42
43
44
45 int main(void)
46 {
47     init();
48     sort(num+1,num+1001,cmp);
49     int icase = 1;
50     int t;scanf("%d",&t);
51     while ( t-- )
52     {
53         int n;scanf("%d",&n);
54         printf("Case %d: ",icase++);
55         printf("%d\n",num[n].id);
56
57     }
58
59
60     return 0;
61 }

代码：