A1023 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

#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;
}