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
 

此題求一個數的2倍是不是這個數的一種排列。

開個陣列記錄0-9的次數，然後對照2倍的數，看看是否完全相同即可。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
char s1[25],s2[25];
int num1[15],num2[15];
int wei=0;
int main()
{
    scanf("%s",s1);
    memset (num1,0,sizeof(num1));
    memset (num2,0,sizeof(num2));
    int len=strlen(s1);
    int jinwei=0,flag=0;
    for (int i=len-1,j=0;i>=0;i--,j++)
    {
        num1[s1[i]-'0']++;
        int t=(s1[i]-'0')*2+jinwei;
        jinwei=t/10;
        s2[j]=t%10+'0';
        num2[s2[j]-'0']++;
    }
    if(jinwei)
    {
        s2[len++]=jinwei+'0';
        num2[s2[len-1]-'0']++;
    }
    for (int i=0;i<10;i++)
    {
        if(num1[i]!=num2[i])
        {
            flag=1;
            break;
        }
    }
    if(flag)
        printf("No\n");
    else
        printf("Yes\n");
    for (int i=len-1;i>=0;i--)
        printf("%c",s2[i]);
    printf("\n");
    return 0;
}