Java的大数计算BigNumber

Posted 2020-10-27 方方的小迷妹

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

不知道为什么有一个样例没过，觉得写得没毛病啊

 1 import java.math.BigInteger;
 2 import java.util.Scanner;
 3 
 4 public class A {
 5 
 6     public static void main(String[] args) {
 7         Scanner cin = new Scanner(System.in);
 8         BigInteger a = cin.nextBigInteger();
 9         BigInteger bb = a;
10         BigInteger c = BigInteger.valueOf(10);
11         BigInteger d;
12         BigInteger ans = a.add(bb);
13         BigInteger b = ans;
14         boolean book = true;
15         int[] book0 = new int[10];
16         while(!a.equals(BigInteger.valueOf(0))) {
17 //            System.out.println(a.equals(0));
18             d = a.mod(c);
19             a = a.divide(c);
20             book0[d.intValue()] = 1;
21         }
22         while(!b.equals(BigInteger.ZERO)) {
23             d = b.mod(c);
24             b = b.divide(c);
25             if(book0[d.intValue()] != 1) {
26                 book = false;
27                 break;
28             }
29         }
30         if(book == false) {
31             System.out.println("No");
32         }
33         else {
34             System.out.println("Yes");
35         }
36         System.out.println(ans);
37 
38     }
39 
40 }