LIGHT OJ 1109 - False Ordering 【因子个数】

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

 

题意：求1-1000内数的因子个数；

思路：1000个数 1000*1000不会超时，直接找因子个数就行（用素数筛的方法和这差不多，优化不了多少），定义结构体，保存数字和该数的因子个数，按照条件排序，预处理一下，直接输出就行；

失误：没有判断i是不是i的因子，虽然没影响，但是还是失误了，没有考虑清楚。

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

struct Node{
	int ord;
	int num;
}stu[1111];

bool cmp(Node a,Node b)
{
	if(a.num!=b.num) 
	return a.num<b.num;
	return a.ord>b.ord;
}

int main()
{
	int i,j,T,N,Kase=0;
	memset(stu,0,sizeof(stu));
	for(i=1;i<=1004;++i)
	{
		stu[i].ord=i;
		for(j=1;j<=i;++j)
		{
			if(i%j==0) stu[i].num++;
		}
	}
	sort(stu+1,stu+1000+1,cmp);
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&N);
		printf("Case %d: %d\n",++Kase,stu[N].ord);
		
	}
	return 0;
}