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 <algorithm>
using namespace std;
int aa[];
int v1[];
int v2[];
int main()
{
string ss;
while(cin>>ss)
{
int i;
for(i=;i<;i++)
{
aa[i]=;
}
for(i=;i<;i++)
{
v1[i]=;
v1[i]=;
}
int count=;
for(i=ss.length()-;i>=;i--)
{
aa[count++]=ss[i]-'';
v1[ss[i]-'']=;
}
for(i=;i<count;i++)
{
aa[i]=aa[i]*;
}
int tem,len;
for(i=;i<count;i++)
{
if(aa[i]>)
{
tem=aa[i]/;
aa[i+]=aa[i+]+tem;
aa[i]=aa[i]%;
}
}
if(aa[count]==) len=count;
else len=count+;
reverse(aa,aa+len);
for(i=;i<len;i++)
{
v2[aa[i]]=;
}
bool ifis=true;
for(i=;i<;i++)
{
if(v1[i]!=v2[i]) ifis=false;
}
if(ifis) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
for(i=;i<len;i++)
{
cout<<aa[i];
}
cout<<endl;
}
return ;
}