Problem:
Score : 100 points

Problem Statement

The door of Snuke's laboratory is locked with a security code.

The security code is a 4-digit number. We say the security code is hard to enter when it contains two consecutive digits that are the same.

You are given the current security code S. If S is hard to enter, print Bad; otherwise, print Good.

Constraints

S is a 4-character string consisting of digits.

Input

Input is given from Standard Input in the following format:

S

Output

If S is hard to enter, print Bad; otherwise, print Good.

Sample Input 1

3776

Sample Output 1

Bad

The second and third digits are the same, so 3776 is hard to enter.

Sample Input 2

8080

Sample Output 2

Good

There are no two consecutive digits that are the same, so 8080 is not hard to enter.

Sample Input 3

1333

Sample Output 3

Bad

Sample Input 4

0024

Sample Output 4

Bad

Solution in C (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>

int quotient(int dividend, int divisor);
int add(int a, int b, int divisor);
int lcm(int a, int b);

int main(void){
    int ans;
    int a,b,c,d;
    scanf("%d %d %d %d", &a,&b,&c,&d);
    ans = b-a+1;
    ans -= add(a, b, c);
    ans -= add(a, b, d);
    ans += add(a, b, lcm(c, d));
    printf("%d", ans);
    return 0;
}

int add(int a, int b, int divisor){
    int ans = 0;
    ans += quotient(b, divisor);
    ans -= quotient(a, divisor);
    if (a % divisor == 0){
        ans += 1;
    }
    return ans;
}

// 商を返す関数
int quotient(int dividend, int divisor){
    int remainder, ans;
    remainder = dividend % divisor;
    ans = (dividend - remainder) / divisor;
    return ans;
}

int lcm(int a, int b){
    int r, x, tmp;
    x = a * b;
 
    /* 自然数 a > b を確認・入替 */
    if(a<b){
    tmp = a;
    a = b;
    b = tmp;
    }

    /* ユークリッドの互除法 */
    r = a % b;
    while(r!=0){
    a = b;
    b = r;
    r = a % b;
    }

    /* 最小公倍数を出力 */
    return x/b;

}