我已经把我朋友的学校任务开始练习了，但我一度都坚持了下来。我认为我的代码是正确的，但如果我看到结果肯定不是。 我的问题如下。程序必须从1到自然数，并找到具有该除数的最小数作为当前数。 例如：[1; 1]，[2; 2]，[3; 4]，[4; 6]，[5; 16]，[6; 12] 因为12是具有6个除数的最小数。 额外的要求是在大约中找到所有100个除数对。标准的5分钟，不是太快，不是太慢的PC。 但是，如果我运行代码，它会停留在第23个数字，并且无法继续下去。 我试图用一个条件缩短操作数（如果当前数字有比现在需要的更多的除数，它会打破循环），但它没有任何区别，我不知道为什么不能找到合适的号码并继续。 如果有人能帮助我，我将不胜感激。 提前谢谢！
public static String message (int numb){
String[] messages = new String[4];
messages[0] = "Working...";
messages[1] = "Done!";
messages[2] = "Please give the interval! [1-100]";
return messages[numb];
}
public static int intervIn (){
Scanner sc = new Scanner(System.in);
int inter = sc.nextInt();
sc.close();
return inter;
}
public static int finding (int need){
int divisor = 1;
int out=1;
if (need!=1){
for (;out<2147483647;){
for (int i = 2;i<=out/2;i++){
if (out%i!=0){
}
else {
divisor++;
if (divisor>=need){
break;
}
}
}
divisor++;
if (divisor==need){
break;
}
else {
divisor=1;
out++;
}
}
}
return out;
}
public static int[][] doit (int row, int column){
int[][] arrayN = new int[row][column];
int divisorNeed = 1;
for (int k = 0;k<row;k++){
arrayN[k][0]=divisorNeed;
arrayN[k][1]=finding(divisorNeed);
divisorNeed++;
}
return arrayN;
}
public static void main(String[] args) {
System.out.println(message(2));
int intervIn = intervIn();
System.out.println(message(0)+'\n');
int[][] arrayNRevis = doit(intervIn,2);
System.out.println(message(1));
for (int i=0;i<intervIn;i++){
System.out.print("[");
for (int j=0;j<2;j++){
System.out.print(arrayNRevis[i][j]+" ");
}
System.out.println("\b]");
}
}
现在输出（差不多8小时后......）：
请给出间隔！ [1-100] 100工作......
[1; 1] [2; 2] [3; 4] [4; 6] [5; 16] [6; 12] [7; 64] [8; 24] [9; 36] [10 ; 48] [11; 1024] [12; 60] [13; 4096] [14; 192] [15; 144] [16; 120] [17; 65536] [18; 180] [19; 262144] [20; 240] [21; 576] [22; 3072]
答案 0 :(得分：0)
每个自然数N可以表示为素数p_i与某些幂k_i的乘法k_i >= 0。所以，假设您的数字n等于：
n = p_1 ^ k_1 * p_2 ^ k_2 * ... * p_z ^ k_z
此号码将有(k_1+1)*(k_2+1)*...*(k_z+1)个分隔符，例如
18 = 2 ^ 1 * 3 ^ 2
并且有（1 + 1）*（2 + 1）个分频器或6个分频器：1,2,3,6,9,18。
素数可以使用筛选算法（https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes）预先计算
然后，您可以使用动态编程来计算给定数字n的分频器数。如果n = n_1 * p_i^k_i（n可以除以p_i^k_i），那么n的除法器数是n_1 *（k_i + 1）的除法器数。
这可以加快分频器计数的计算。
答案 1 :(得分：0)
显然你并没有停留在第23个号码上，但需要大约120个小时来评估它。见http://oeis.org/A005179
这是一个APL NARS2000工作区，可以大大加快速度。 前100只需要1/10秒。
)WSID
C:\Users\Ab\AppData\Roaming\NARS2000\workspaces\combfactor
)FNS
combfactor factors listws save
)VARS
COMBFACTOR DESCRIBE DIV RESULTSfor100
∇ z ← combfactor n;f;p
[1] f ← (⊂,n),factors n ⋄ ⍝ find all factors combinations of n
[2] p ← ⍸0π⍳¯2π32 ⋄ ⍝ The first 32 primes numbers
[3] z ← ((⍴¨f)↑¨⊂⍟p)+.רf-1 ⋄ ⍝ give ratios of all combinations
[4] z ← ∊¯1+((⌊/z)=z)/f ⋄ ⍝ get the combination with minimum ratio
[5] z ← (0x+(⍴z)↑p)×.*z ⋄ ⍝ evaluate p_1^(f_1-1) * p_2^(f_2-1) * ... * p_n^(f_n-1)
∇
∇ z ← {fmax} factors n;f;d
[1] :if 0 = ⎕NC 'fmax' ⋄ fmax ← 2*63 ⋄ :endif
[2] z ← ⍬ ⋄ f ← ⌊fmax⌊n÷2
[3] :while f ≥ 2
[4] :if 0 = f∣n
[5] d ← n÷f ⋄ :if d ∧.≤ f,fmax ⋄ z ,← ⊂f,d ⋄ :endif
[6] z ,← f,¨f factors d
[7] :endif ⋄ f -← 1
[8] :endwhile
∇
COMBFACTOR
RESULTSfor100 ← ⍪(,¨⍳100),¨combfactor¨,¨⍳100 ⋄ ⍝ 0.1 second
DESCRIBE
DESCRIBE
⍝ def factors(number, max_factor=sys.maxint):
⍝ result = []
⍝
⍝ factor = min(number / 2, max_factor)
⍝ while factor >= 2:
⍝ if number % factor == 0:
⍝ divisor = number / factor
⍝
⍝ if divisor <= factor and divisor <= max_factor:
⍝ result.append([factor, divisor])
⍝
⍝ result.extend([factor] + item for item in factors(divisor, factor))
⍝
⍝ factor -= 1
⍝
⍝ return result
⍝
⍝ print factors(12) # -> [[6, 2], [4, 3], [3, 2, 2]]
⍝ print factors(24) # -> [[12, 2], [8, 3], [6, 4], [6, 2, 2], [4, 3, 2], [3, 2, 2, 2]]
⍝ print factors(157) # -> []
DIV
divisors←{z[⍋z←∊∘.×/1,¨(∪π⍵)*¨⍳¨∪⍦π⍵]}
RESULTSfor100
1 1
2 2
3 4
4 6
5 16
6 12
7 64
8 24
9 36
10 48
11 1024
12 60
13 4096
14 192
15 144
16 120
17 65536
18 180
19 262144
20 240
21 576
22 3072
23 4194304
24 360
25 1296
26 12288
27 900
28 960
29 268435456
30 720
31 1073741824
32 840
33 9216
34 196608
35 5184
36 1260
37 68719476736
38 786432
39 36864
40 1680
41 1099511627776
42 2880
43 4398046511104
44 15360
45 3600
46 12582912
47 70368744177664
48 2520
49 46656
50 6480
51 589824
52 61440
53 4503599627370496
54 6300
55 82944
56 6720
57 2359296
58 805306368
59 288230376151711744
60 5040
61 1152921504606846976
62 3221225472
63 14400
64 7560
65 331776
66 46080
67 73786976294838206464
68 983040
69 37748736
70 25920
71 1180591620717411303424
72 10080
73 4722366482869645213696
74 206158430208
75 32400
76 3932160
77 746496
78 184320
79 302231454903657293676544
80 15120
81 44100
82 3298534883328
83 4835703278458516698824704
84 20160
85 5308416
86 13194139533312
87 2415919104
88 107520
89 309485009821345068724781056
90 25200
91 2985984
92 62914560
93 9663676416
94 211106232532992
95 21233664
96 27720
97 79228162514264337593543950336
98 233280
99 230400
100 45360
这是一个C程序， 62微秒！
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define min(a,b) (((a) < (b)) ? (a) : (b))
#define MaxInt 0x7fffffffffffffff
#define ll long long
int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,
67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131};
ll minimums[16]; // contains minimum for each level
ll ipow(ll base, int exp) {
ll result = 1;
while (exp) {
if (exp & 1) result *= base;
exp >>= 1; base *= base;
} return result;
}
ll factors(ll number, ll max_factor, int level) {
ll result = MaxInt;
ll factor = min(number / 2, max_factor);
while (factor >= 2) {
if (number % factor == 0) {
ll divisor = number / factor;
ll tempf = ipow(primes[level], factor-1);
if (divisor <= factor && divisor <= max_factor) {
ll tempd = ipow(primes[level+1], divisor-1);
minimums[level] = min(minimums[level], tempf * tempd);
}
ll fct = factors(divisor, factor, level+1);
if (fct < MaxInt)
minimums[level] = min(minimums[level], tempf * fct);
result = minimums[level];
minimums[level+1] = MaxInt;
}
factor -= 1;
}
return result;
}
ll fact(int number) {
for (int level = 0; level < 16; level++) minimums[level] = MaxInt;
ll res = factors(number, MaxInt, 0);
if (res < MaxInt) return res;
else return 0;
}
int main(int argc, char *argv[]) {
int N = 100;
if (argc > 1) N = atoi(argv[1]);
ll res[N];
clock_t Start = clock();
for(int n = 1; n <= 10000; n++)
for (int i = 1; i <= N; i++) res[i] = fact(i);
printf("\n%0.6f second(s)\n", (clock() - Start)/1000.0/10000.0);
for (int i = 1; i <= N; i++)
if (res[i] > 0) printf("%d %lld\n", i, res[i]);
else if (i > 64 ) printf("%d 2 power %d\n", i, i-1);
else printf("%d %lld\n", i, ipow(2, i-1));
return 0;
}