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:

# include<cstdio>
# include<iostream>
# include<algorithm>

using namespace std;

# define MAX 1234

int a[MAX];

struct node
{
    int val;
    int id;
}num[MAX];


int cmp ( const struct node & x,const struct node & y )
{
    if ( x.val==y.val )
    {
        return x.id > y.id;
    }
    return x.val < y.val;
}


void init()
{
    for ( int i = 1;i <= 1000;i++ )
    {
        num[i].id = i;
        for ( int j = 1;j <= i;j++ )
        {
            if ( i%j == 0 )
            {
                num[i].val++;
            }
        }
    }
}



int main(void)
{
    init();
    sort(num+1,num+1001,cmp);
    int icase = 1;
    int t;scanf("%d",&t);
    while ( t-- )
    {
        int n;scanf("%d",&n);
        printf("Case %d: ",icase++);
        printf("%d\n",num[n].id);

    }


    return 0;
}

 

 

代码：