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");
        }
    }
}