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 <cstdio>
#include <stdlib.h>
#include <cstring>
#include <iostream>
#include <math.h>
using namespace std;
struct bign{
int d[21];
int len;
bign(){
memset(d,0,sizeof(d));
len=0;
}
};
bign change(char str[]){
bign a;
a.len=strlen(str);
for(int i=0;i<a.len;i++){
a.d[i]=str[a.len-i-1]-'0';
}
return a;
}
bign multi(bign a,int b){
bign c;
int carry=0;//进位
for(int i=0;i<a.len;i++){
int temp=a.d[i]*b+carry;
c.d[c.len++]=temp%10;
carry=temp/10;
}
while(carry!=0){
c.d[c.len++]=carry%10;
carry/=10;
}
return c;
}
bool judge(bign a,bign b){
if(a.len!=b.len) return false;
int count[10]={0};
for(int i=0;i<a.len;i++){
count[a.d[i]]++;
count[b.d[i]]--;
}
for(int i=0;i<10;i++){
if(count[i]!=0)
return false;
}
return true;
}
void print(bign a){
for(int i=a.len-1;i>=0;i--)
printf("%d",a.d[i]);
}
int main(){
char str[21];
gets(str);
bign a=change(str);
bign mul=multi(a,2);
if(judge(a,mul)) printf("Yes\n");
else printf("No\n");
print(mul);
return 0;
}
