1023. Have Fun with Numbers (20)

最新推荐文章于 2022-11-24 21:02:34 发布

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最新推荐文章于 2022-11-24 21:02:34 发布

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本文介绍了一个有趣的数字处理问题：判断一个数翻倍后是否仅由原数位的排列组成。通过比较两个字符串中各字符出现次数的方法来解决这个问题，并提供了一段实现该功能的C++代码。

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

提交代码

直接记录比较两个字符串中每个字符出现的次数就可以了
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
using namespace std;
char str[30];
char str1[30];
int book[10] ={0};
int book1[10] ={0};
int main(){
//  freopen("input.txt","r",stdin);
  scanf("%s",str);
  int a;
  int jin = 0; 
  int len = strlen(str);
  for(int i = 0;i<len;i++){
    book[str[i]-'0']++;
  }
  for(int i = len-1;i >= 0;i--){
    a = (str[i]-'0')*2+jin;
    jin = a/10;
    str1[i] = a%10+'0';
  }
  int flag = 1;
  if(jin != 0){
    
    printf("No\n");
    printf("1");
    printf("%s\n",str1);
  } else{
    for(int i = 0;i< len;i++){
      book1[str1[i]-'0']++;
    }
    for(int i = 0;i < 10;i++){
      if(book1[i] != book[i]){
        flag = 0;
      }
    }
    if(flag == 0){
      printf("No\n");
      printf("%s\n",str1);
    }else{
      printf("Yes\n");
      printf("%s\n",str1);
    }
  }
  return 0;
}