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 }
原文地址：https://www.cnblogs.com/xuyanqd/p/8590785.html