Java 生成序列/返回序列中的第n个项

我需要生成一个序列，使其成员只包含

1

，

2

，

3

位。例如，

12311121321231333111….

等等，直到

10^18

术语。 我无法推断出这方面的任何模式。似乎不可能在系列中编写最多10^18个术语的代码

1、2、3、11、12、13、21、22、23、31、32、33、111、112、113、121、122、， 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333, 1111, 1112, 1113, 1121, 1122, 112311311321133211211211211212121212122121

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我希望在序列中找到给定的第n项。这是一种数字系统，它只包含

1

、

2

、

3

，或者是这些数字的组合，就像我们通常的数字系统一样，按顺序解释为一个数字。

这只是一种基数为3的数字系统，只是数字从1到3，而不是从0到2。数学的计算方法是一样的：

1=1*3^0
2=2*3^0
3=3*3^0
4=1*3^1+1*3^0
5=1*3^1+2*3^0
6=1*3^1+3*3^0
7=2*3^1+1*3^0
…
19=1*3^2+3*3^1+1*3^0

写两种方法：

数字（n）：为给定的

n

计算最右边的数字。一些测试用例：数字（4）=1，数字（5）=2，数字（15）=3

剩余（n）：计算表示n但最右边数字被截断的数字。一些测试用例：遗留（4）=1，遗留（15）=4，遗留（23）=7

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现在，将这两种方法结合到问题的解决方案中，反复切掉最右边的数字，直到一无所有。您可能会发现递归执行此操作更容易。

您已经提到的序列称为。它是由

您可以在上面的链接中查看公式的解释。到目前为止，我已经使用

long

编写了它的基本级别。要达到

10^18

术语，您需要使用Java类

class SequenceGeneratorWith123 {

    // Written by Soner

    private static double logOfBase(long base, long num) {
        return Math.log(num) / Math.log(base);
    }

    private static int mfunc(long n) {
        return (int) Math.floor(logOfBase(3, 2 * n + 1));
    }

    private static int b(int j, double m, long n) {
        return (int) Math.floor((2 * n + 1 - Math.pow(3, m)) / (2 * Math.pow(3, j)));
    }

    public static void main(String[] args) {

        for (int i = 0; i < 9; i++) {
            long n = (long) Math.pow(10, i);
            int m = mfunc(n);
            long sum = 0;

            for (int j = 0; j < m ; j++) {
                sum += ((1 + b(j, m, n) % 3) * Math.pow(10, j));
            }
            System.out.printf("a(10^%d) = %d\n", i, sum);
        }

        System.out.println("After the point, overflow will occur " +
                        "because of long type.");    
    }
}

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您只需要玩一下代码，也就是说，我们只需稍微更改一下

main（）

，就可以实现您的愿望

long n = 1;
// How many terms you need you can alter it by pow() method.
// In this example 10^2 = 100 terms will be obtained.
int term = (int)Math.pow(10, 2);
for (int i = 0; i < term; i++) {

    int m = mfunc(n);
    long sum = 0;

    for (int j = 0; j < m ; j++) {
        sum += ((1 + b(j, m, n) % 3) * Math.pow(10, j));
    }
    System.out.printf("%d. term = %d\n", i + 1, sum);
    n++;
}

---

序列中有多少个长度为k的项目？由此，您可以编写一个有效的程序来确定序列中第n项的长度吗？是的，对于同一件事，我推导了这么多术语，但在这之后，我感到困惑。你能为我简化一下吗？因为我是编程新手，所以N最多可以是10^18个项，所以我认为如果我们在输入的同时计算它，在竞争性编程中可能不会那么有效。

a(10^0) = 1
a(10^1) = 31
a(10^2) = 3131
a(10^3) = 323231
a(10^4) = 111123331
a(10^5) = 11231311131
a(10^6) = 1212133131231
a(10^7) = 123133223331331
a(10^8) = 13221311111312132
After the point, overflow will occur because of long type.

long n = 1;
// How many terms you need you can alter it by pow() method.
// In this example 10^2 = 100 terms will be obtained.
int term = (int)Math.pow(10, 2);
for (int i = 0; i < term; i++) {

    int m = mfunc(n);
    long sum = 0;

    for (int j = 0; j < m ; j++) {
        sum += ((1 + b(j, m, n) % 3) * Math.pow(10, j));
    }
    System.out.printf("%d. term = %d\n", i + 1, sum);
    n++;
}

1. term = 1
2. term = 2
3. term = 3
4. term = 11
5. term = 12
6. term = 13
7. term = 21
8. term = 22
9. term = 23
10. term = 31
11. term = 32
12. term = 33
13. term = 111
14. term = 112
15. term = 113
16. term = 121
17. term = 122
18. term = 123
19. term = 131
20. term = 132
21. term = 133
22. term = 211
23. term = 212
24. term = 213
25. term = 221
26. term = 222
27. term = 223
28. term = 231
29. term = 232
30. term = 233
31. term = 311
32. term = 312
33. term = 313
34. term = 321
35. term = 322
36. term = 323
37. term = 331
38. term = 332
39. term = 333
40. term = 1111
41. term = 1112
42. term = 1113
43. term = 1121
44. term = 1122
45. term = 1123
46. term = 1131
47. term = 1132
48. term = 1133
49. term = 1211
50. term = 1212
51. term = 1213
52. term = 1221
53. term = 1222
54. term = 1223
55. term = 1231
56. term = 1232
57. term = 1233
58. term = 1311
59. term = 1312
60. term = 1313
61. term = 1321
62. term = 1322
63. term = 1323
64. term = 1331
65. term = 1332
66. term = 1333
67. term = 2111
68. term = 2112
69. term = 2113
70. term = 2121
71. term = 2122
72. term = 2123
73. term = 2131
74. term = 2132
75. term = 2133
76. term = 2211
77. term = 2212
78. term = 2213
79. term = 2221
80. term = 2222
81. term = 2223
82. term = 2231
83. term = 2232
84. term = 2233
85. term = 2311
86. term = 2312
87. term = 2313
88. term = 2321
89. term = 2322
90. term = 2323
91. term = 2331
92. term = 2332
93. term = 2333
94. term = 3111
95. term = 3112
96. term = 3113
97. term = 3121
98. term = 3122
99. term = 3123
100. term = 3131