PAT A 1023 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 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

大数运算加HashTable

#include<cstdio>
#include<cstring>

struct bign {
    int d[25];
    int len;
    bign() {
        memset(d, 0, sizeof(d));
        len = 0;
    }
};

int Hashtable1[10] = { 0 }, Hashtable2[10] = { 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';
        Hashtable1[a.d[i]]++;
    }
    return a;
}

bign Double(bign a) {
    bign b;
    b.len = a.len;//先令长度相等，不等也就差一位
    int carry = 0, temp;
    for (int i = 0; i < a.len; i++) {
        temp = a.d[i] * 2 + carry;
        b.d[i] = temp % 10;
        carry = temp / 10;
        Hashtable2[b.d[i]]++;
    }
    if (carry) {
        b.len++;
        b.d[b.len - 1] = carry;
    }
    return b;
}

int main() {
    char s[25];
    scanf("%s", s);
    bign a, b;
    a = change(s);
    b = Double(a);
    int i;
    for ( i = 0; i < 10; i++) {
        if (Hashtable1[i] != Hashtable2[i]) {
            printf("No\n");
            break;
        }
    }
    if (i == 10) printf("Yes\n");
    for (i = b.len - 1; i >= 0; i--) printf("%d", b.d[i]);
    return 0;
}