PAT (Advanced Level) Practise 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

代码：

#include <cstdio>
#include <cstring>
struct bign{
    int d[21];
    int len;
    bign(){
        memset(d,0,sizeof(d));
        len=0;
    }
};

bign change(char s[]){
    bign a;
    a.len=strlen(s);
    for(int i=0;i<a.len;i++) a.d[i]=s[a.len-1-i]-'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){
        c.d[c.len++]=carry%10;
        carry/=10;
    }
    return c;
}

int judge(bign a,bign b){
    if(a.len!=b.len) return 0;
    int cnt[10]={0};
    for(int i=0;i<a.len;i++){
        cnt[a.d[i]]++;
        cnt[b.d[i]]--;
    }
    for(int i=0;i<10;i++){
        if(cnt[i]) return 0;
    }
    return 1;
}

void print(bign a){
    for(int i=a.len-1;i>=0;i--) printf("%d",a.d[i]);
}

int main(){
    char s[21];
    gets(s);
    bign a=change(s);
    bign mul=multi(a,2);
    if(judge(a,mul)) printf("Yes\n");
    else printf("No\n");
    print(mul);
    return 0;
}