LightOJ 1109 - False Ordering (因式分解)

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 n^th 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：