1023. Have Fun with Numbers (20)

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<cstdio>
#include<cstring>
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;
    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) == true) printf("Yes\n");
    else printf("No\n");
    print(mul);
    return 0;
}