Have Fun with Numbers (玩数字) java

最新推荐文章于 2021-05-22 12:19:52 发布

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本文探讨了一种独特的数学现象，即某个特定数字在加倍后，其结果仅由原数字的位数重新排列组成，且不含重复。文章通过示例说明了这一特性，并提供了一个Java程序，用于检查任意给定的不超过20位的大数字是否具有这种特殊的倍增性质。

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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 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

/*数字比较大,要用BigInteger来完成*/
import java.math.BigInteger;
import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		int[] count1 = new int[10];
		int[] count2 = new int[10];
		Scanner sc=new Scanner(System.in);
		String str=sc.next();
		sc.close();
		BigInteger bi=new BigInteger(str);
		bi=bi.multiply(new BigInteger("2"));
		String newstr=bi.toString();
		if(newstr.length()!=str.length()) {
			System.out.println("No");
			System.out.println(newstr);
			return;
		}
		for(int i=0;i<str.length();i++) {
			count1[str.charAt(i)-'0']++;
		}
		for(int i=0;i<newstr.length();i++) {
			count2[newstr.charAt(i)-'0']++;
		}
		for(int i=0;i<10;i++) {
			if(count1[i]!=count2[i]) {
				System.out.println("No");
				System.out.println(newstr);
				return ;
			}
		}
		System.out.println("Yes");
		System.out.println(newstr);
	}
}