数据结构练习 00-自测4. Have Fun with Numbers (20)

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>
#include <sstream>
using namespace std;
int sort(int a[],int n){
    int temp;
    for(int i=0;i<n;i++){
        for(int j=i;j<n;j++){
            if(a[i]>a[j]){
                temp=a[i];
                a[i]=a[j];
                a[j]=temp;
            }
        }
    }
    return 0;
}
int main(){
    string num;
    stringstream ss;
    int size,j=0;
    cin>>num;
    size=num.size();
    int *a=new int[size];
    int *b=new int[size+1];
    int *doubleNum=new int[size+1];
    for(int i=0;i<size+1;i++){
        doubleNum[i]=0;
        b[i]=0;
    }
    for(int i=0;i<size;i++){
        a[i]=num[i]-48;
    }
    for(int i=size;i>0;i--){
        if(a[i-1]+a[i-1]>=10){
            doubleNum[i]+=(a[i-1]+a[i-1])%10;
            doubleNum[i-1]+=1;
        }else{
            doubleNum[i]+=a[i-1]+a[i-1];
        }
    }
    if(doubleNum[0]==0){
        for(int i=0;i<size;i++){
            b[i+1]=doubleNum[i+1];
        }
        sort(doubleNum,size+1);
        sort(a,size);
        for(int i=0;i<size;i++){
        if(a[i]==doubleNum[i+1]){
            j++;
        }
        }
        if(j==size){
            cout<<"Yes"<<endl;
            for(int i=0;i<size;i++){
                cout<<b[i+1];
            }

        }else{
            cout<<"No"<<endl;
            for(int i=0;i<size;i++){
                cout<<b[i+1];
            }

        }
    }else{
        cout<<"No"<<endl;
        for(int i=0;i<size+1;i++){
            cout<<doubleNum[i];
        }
    }

}

测试结果