(Q-3-2) DNA ratio of 'b' and 'c'

#include <stdio.h>

double ratio(char arr[])
{
    int b=0,c=0;
    double ratio;
    for(int i=0;i<100;i++)
        {
            if(arr[i]=='c')
                c++;
            else if(arr[i]=='b')
                b++;
        }
    if(c==0)
        return -1;
    else
        ratio=(double)b/c;

return ratio;


}


int main()
{
    char DNA[100];
    printf("Input: ");
    gets(DNA);

    double result=ratio(DNA);
    if(result==-1)
    {
        printf("All characters are 'b'");
         return 0;
    }
    else
        printf("Ration of 'b' to 'c' is: %.3lf", result);

    return 0;
}

&&

#include <stdio.h>
void ratio(char str[])
{
    int count1 = 0, count2 = 0, x, y;

    for (int i = 0; i < 100; i++)
    {
        if (str[i] == 'b')
        {
            count1++;
        }
    }
    for (int i = 0; i < 100; i++)
    {
        if (str[i] == 'c')
        {
            count2++;
        }
    }

    float r;
    r = count1 / count2;

    printf("Ratio b and c is %d : %d or %.2f", count1, count2, r);
}

int main()
{
    char DNA[100];

    printf("Enter random numbers of b and c:\n");
    scanf("%s", DNA);

    ratio(DNA);

    return 0;
}