1023 Have Fun with Numbers

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 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

思路：
　　1、求2倍可以认为是两个数相加
　　2、首先检查加倍后数的位数是否相同，如果不相同则一定不是
　　3、如果位数相同，我的思路是使用dfs对原数进行深搜，本质是使用了dfs求全排列
　　4、其他人的思路是判断原数和加倍后的每个数中出现的0-9的个数是否相同（比我的思路好多了，虽然我的也能ac）
　　

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>v1;
vector<int>v2;
int visited[23];
void doubleNum()
{
    int temp=0;//保存进位
    for(auto num:v1)
    {
        int a=num*2+temp;
        if(a>=10)
        {
            temp=1;
            v2.push_back(a-10);
        }
        else
        {
            temp=0;
            v2.push_back(a);
        }
    }
    if(temp>0)
        v2.push_back(temp);
    reverse(v2.begin(),v2.end());
}

void output()
{
        for(auto num:v2)
            cout<<num;
        cout<<endl;
       // cout<<v2[0]<<endl;
}
bool flag=false;

void dfs(int i,int n,int current)
{
    //cout<<v1[i]<<" "<<v2[current]<<endl;
    if(v1[i]!=v2[current])
        return;
    if(current==n-1)
    {
        flag=true;
        return;

    }
    for(int j=0;j<n;j++)
    {
         if(visited[j]!=1)
         {
             visited[j]=1;
             dfs(j,n,current+1);
             visited[j]=0;
         }
    }
    //return false;
}

int main()
{
    string num;
    cin>>num;
    v1.resize(num.size());
    for(int i=0;i<num.size();i++)
        v1[num.size()-1-i]=num[i]-'0';
    //set<int>s;
    doubleNum();

    if(v1.size()!=v2.size())
    {
        cout<<"No"<<endl;
        output();
    }
    else
    {
        int n=num.size();
        int i;
        for(i=0;i<n;i++)
        {
            if(v1[i]==v2[0])
                break;
        }
        //cout<<i<<endl;
        visited[i]=1;
        if(i<n)
            dfs(i,n,0);
        if(flag)
        {
            cout<<"Yes"<<endl;
            output();
        }
        else
        {
            cout<<"No"<<endl;
            output();
        }
    }
    return 0;
}