False Ordering（统计因子个数、素因子分解）

最新推荐文章于 2022-01-02 13:38:31 发布

转载最新推荐文章于 2022-01-02 13:38:31 发布 · 155 阅读

本文探讨了如何通过计算整数的因子数量来进行特定排序，并提供了优化算法来解决实际问题。

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False Ordering

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1109

Description

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

1000

Sample Output

Case 1: 1

Case 2: 997

Case 3: 991

Case 4: 983

Case 5: 840

统计一个整数n的因子个数的方法：将n进行素因子分解，n=(q1^r1)*(q2^r2)*...*(qi^ri)，则n的因子数为（r1+1）*(r2+1)*...*(ri+1).

因为对于每个q，其对应的指数为r，q可取0个，1个，...，r-1个。

AC Code：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 int dr[1001], num[1001];  //dr[i]是i的因子个数，num[i] = i
 8 int d[500], di = 0;
 9 bool p[1001];
10 
11 bool cmp(const int& a, const int& b)
12 {
13     if(dr[a] != dr[b]) return dr[a] < dr[b];
14     return  a > b;
15 }
16 
17 void Init_Prime()
18 {
19     memset(p, 0, sizeof(p));
20     p[1] = 1;
21     for(int i = 2; i < 1001; i++)
22     {
23         if(!p[i])
24         {
25             d[di++] = i;
26             for(int j = 2; j * i < 1001; j++)
27                 p[i*j] = 1;
28         }
29     }
30     //for(int i = 0; i < di; i++) cout << d[i] << ' ';
31     return ;
32 }
33 
34 void Cal(int x)
35 {
36     int t = x;
37     int sum = 1, cnt;
38     for(int i = 0; d[i] <= t / 2 && i < di; i++)
39     {
40         cnt = 0;
41         while(x % d[i] == 0)
42         {
43             cnt++;
44             x /= d[i];
45         }
46         sum *= (cnt + 1);
47     }
48     if(!p[t]) sum *= 2;
49     dr[t] = sum;
50     return ;
51 }
52 
53 int main()
54 {
55     int t, n, ca = 1;
56     scanf("%d", &t);
57     Init_Prime();
58     for(int i = 1; i < 1001; i++)
59     {
60         num[i] = i;
61         Cal(i);
62     }
63     //for(int i = 1; i < 30; i++) cout << i << ' ' << dr[i] << endl;
64     sort(num + 1, num + 1001, cmp);
65     while(t--)
66     {
67         scanf("%d", &n);
68         printf("Case %d: %d\n", ca++, num[n]);
69     }
70     return 0;
71 }