1023 Have Fun with Numbers (20 分) java 题解

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

解题思路：

题目大意：给定一个大数，判断翻倍后的数在不考虑位置时与其是否有相同的数字和出现次数。

大数运算调用BigInteger类的方法，判断出现次数可以用数组模拟哈希表，前后运算检验最后数组是否都为0。

java代码：

import java.io.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String str = br.readLine();
		BigInteger bignum = new BigInteger(str);
		BigInteger multiply = bignum.multiply(new BigInteger("2"));
		int []arr = new int[10];
		String s = multiply + "";
		for(int i = 0; i < s.length();i++) {
			arr[s.charAt(i) - '0']++;
		}
		for(int i = 0; i < str.length();i++) {
			arr[str.charAt(i) - '0']--;
		}
		for(int i = 0; i < arr.length;i++) {
			if(arr[i] != 0) {
				System.out.println("No");
				System.out.println(s);
				System.exit(0);
			}
		}
		System.out.println("Yes");
		System.out.println(s);
	}
}

PAT提交截图：