1023. Have Fun with Numbers (20)-优快云博客

本文介绍了一种特殊的数字性质：当一个由1至9组成的9位数翻倍后，其结果依然是这些数字的一种排列。文章提供了一个算法，用于判断任意给定的不超过20位的正整数，在翻倍后是否仅由原数位的不同排列组成。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string s;
    int head;
    cin>>s;
    int *arr = new int[s.length()];
    int *doubleArr = new int[s.length()];
    for(int i=0;i<s.length();i++){
        arr[i] = s[i]-'0';
    }
    head = arr[0];
    int flag = 0;
    for(int i=s.length()-1;i>=0;i--){
        if(flag==0){
            if(arr[i]*2>=10){
                flag = 1;
                doubleArr[i] = (arr[i]*2)%10;
            }else{
                doubleArr[i] = (arr[i]*2)%10;
                flag = 0;
            }
        }else{
            if((arr[i]*2+1)>=10){
                flag = 1;
                doubleArr[i] = (arr[i]*2+1)%10;
            }else{
                doubleArr[i] = (arr[i]*2+1)%10;
                flag = 0;
            }
        }
    }
    int isFunNum = 0;
    for(int i=0;i<s.length();i++){
        int findL = 0;
        for(int j=0;j<s.length();j++){
            if(doubleArr[i]==arr[j]){
                findL = 1;
                arr[j] = -1;
                break;
            }
        }
        if(findL==0){
            isFunNum = 0;
            break;
        }
        isFunNum = 1;
    }
    if(isFunNum == 0){
        cout<<"No"<<endl;
        if(doubleArr[0]<=head)
            cout<<1;
        for(int i=0;i<s.length();i++){
            cout<<doubleArr[i];
        }
    }else{
        cout<<"Yes"<<endl;
        if(doubleArr[0]<=head)
            cout<<1;
        for(int i=0;i<s.length();i++){
            cout<<doubleArr[i];
        }
    }
}