PAT 1023 Have Fun with Numbers

1023 Have Fun with Numbers （20 分)

 

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

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define maxnum 100005

int a[] = {};
int count1[] = {}; //分别对倍乘前后的位计数
int count2[] = {};

int main(){
    string s;
    cin >> s;
    int len = s.size();
    for(int i=;i < s.size();i++){
        a[i] = s[i]-'';
    }
    for(int i=;i < len;i++){
        count1[a[i]]++;
    }
    for(int i=;i < len/;i++){ //翻转
        swap(a[i],a[len-i-]);
    }

    int jinwei = ;   //倍乘
    for(int i=;i <= len;i++){
        int num = a[i]* + jinwei;
        a[i] = num%;
        jinwei = (num)/;
    }

//    for(auto num:a) cout << num << " ";

    int pos = ;
    for(int i=;i >= ;i--){
        if(a[i]){pos = i;break;}
    }
    for(int i=;i <= pos;i++){
        count2[a[i]]++;
    }

    int flag = ;
    for(int i=;i < ;i++){
        if(count1[i] != count2[i])flag = ;
    }
    if(flag) cout << "Yes" << endl;
    else cout << "No" << endl;

    for(int i=pos;i >= ;i--){
        cout << a[i];
    }

    return ;
}

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