2018 ACM-ICPC 中国大学生程序设计竞赛线上赛 I. Reversion Count (java大数)

Description:

There is a positive integer X, X's reversion count is Y. For example, X=123, Y=321; X=1234, Y=4321. Z=(X-Y)/9, Judge if Z is made up of only one number(0,1,2...9), like Z=11,Z=111,Z=222,don't consider '+'and '-'.

Input:

Input contains of several test cases. Each test case only contains of a number X, L is the length of X. ( 2 <= L < 100)

Output:

Output “YES”or “NO”.

样例输入

10
13

样例输出

YES
YES

题意：一个数与自己的翻转做差，最后结果的数是不是只有一个数字组成
思路：利用java中的BigInteger与StringBuffer类
代码如下：

 import java.math.*;
 import java.security.Principal;
 import java.util.*;

 import javax.xml.bind.helpers.AbstractMarshallerImpl;

 import java.*;
 public class Main {

     public static void main(String[] args) {
         Scanner cin = new Scanner(System.in);
         BigInteger x;
         String xx,yyy;
         int cnt[] = new int[10];
         while (cin.hasNext()){
             for (int i=0;i<10;i++)
                 cnt[i]=0;
             xx = cin.nextLine();
             StringBuffer yy = new StringBuffer(xx).reverse();
             yyy = yy.toString();
             BigInteger a = new BigInteger(xx);
             BigInteger b = new BigInteger(yyy);
             if (a.compareTo(b)>=0){
                 a = a.subtract(b);
             }
             else a = b.subtract(a);
             a = a.divide(BigInteger.valueOf(9));
             String s = a.toString();
             for (int i=0;i<s.length();++i){
                 cnt[s.charAt(i)-'0' ]++;
             }
             //System.out.println(a);
             int ans = 0;
             for (int i=0;i<10;i++){
                 if (cnt[i]!=0) ans++;
             }
             if (ans==1) System.out.println("YES");
             else System.out.println("NO");
         }
     }
 }