1023. Have Fun with Numbers

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    Scanner in=new Scanner(System.in);
    int init[]={0,0,0,0,0,0,0,0,0,0};
    String s=in.nextLine();
    int length = s.length();
    int[] str1 = new int[length];
    for(int i = 0; i < length; i++) {
      str1[i] = s.charAt(i) - '0';
    }
    int[] str2 =new int[21];
    int add=0,i=0, num=0,tmp=0;
    for (int k:str1){
        init[k]++;
      }
    for(int g=str1.length-1;g>=0;g--){
      tmp=str1[g];
      num=tmp*2+add;
      if(num>=10){
        str2[i++]=num%10;
        add=num/10;
      }
      else{
        str2[i++]=num;
        add=0;
      }
    }
    if(add!=0){
      str2[i++]=add;
    }
    for(int k=i-1;k>=0;k--){
      init[str2[k]]--;
    }
    int flag=0;
    for (int k=0;k<10;k++){
      if(init[k]!=0) {
        flag=1;
        break;
      }
    }
    if(flag==1) System.out.println("No");
    else System.out.println("Yes");
    for (int k=i-1;k>=0;k--){
      System.out.print(str2[k]);
    }
    System.out.println();
  }

}

这题要注意的是：

1.最高位如果有进位不要漏掉；

2.字符串的处理吧，之前写的代码用的split一直通过不了，不知道为啥在同学的机子上跑出来的话最高位会缺失，但是我自己的就正常可能是这里有些问题吧，不懂，希望之后能想明白orz

错误的代码如下

import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    Scanner in=new Scanner(System.in);
    int init[]={0,0,0,0,0,0,0,0,0,0};
    String s=in.nextLine();
    String str1[]=s.split("");
    int[] str2 =new int[21];
    int add=0,i=0, num=0,tmp=0;
    for (String k:str1){
      init[Integer.parseInt(k)]++;
    }
    for(int g=str1.length-1;g>=0;g--){
      tmp=Integer.parseInt(str1[g]);
      num=tmp*2+add;
      if(num>=10){
        str2[i++]=num%10;
        add=num/10;
      }
      else{
        str2[i++]=num;
        add=0;
      }
    }
    if(add!=0){
      str2[i++]=add;
    }
    for(int k=i-1;k>=0;k--){
      init[str2[k]]--;
    }
    int flag=0;
    for (int k=0;k<10;k++){
      if(init[k]!=0) {
        flag=1;
        break;
      }
    }
    if(flag==1) System.out.println("No");
    else System.out.println("Yes");
    for (int k=i-1;k>=0;k--){
      System.out.print(str2[k]);
    }
    System.out.println();
  }

}