LOJ 1109 - False Ordering （排序）

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本文介绍了一道关于整数因子个数排序的算法题目。通过结构体排序与预处理表格的方法，实现对1到1000之间的整数按其因子数量进行排序，并输出特定位置的数值。

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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 n^th number after ordering.

Sample Input

Output for Sample Input

1000

Case 1: 1

Case 2: 997

Case 3: 991

Case 4: 983

Case 5: 840

题意：将1到1000的所有数按照他的因子个数从小到大排列，如果有相同的因子个数，则最大的那个数排在前面，然呢输出第n个数

思路：结构体排序加打表

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
struct Node{
	int x,y;
	int ans;
}node[1010];
bool cmp(Node a,Node b)
{
	if(a.ans == b.ans)
	   return a.x > b.x;
	else
	   return a.ans < b.ans;
}
void dabiao()
{
	for(int i = 1 ; i <= 1000 ; i++)
	{
		node[i].x = i;
		node[i].y = i;
	}
	node[1].ans = 1;
	for(int i = 2 ; i <= 1000 ; i++){
		int k = sqrt(i);
		if(k * k == i)
	   	      node[i].ans--;
		for(int j = 1 ; j <= k; j++)
	    {
	   	   if(node[i].y % j == 0)
	   	      node[i].ans += 2; 	   
	    }
	 
	}
}
int main()
{
	int t,kcase = 1;
	dabiao();
	sort(node+1,node+1001,cmp);
	int n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
	    printf("Case %d: %d\n",kcase++,node[n].x);
	}
	return 0;
}