Nyoj 949 Harry Potter (detail question)

Harry Potter time limit: +Ms | Memory Limit:65535KB Difficulty:3

Descriptive narrative

Harry learned three kinds of magic. He was able to use the first magic to turn a gram of sand into a B-gram metal, to use another kind of magic to turn C-gram metal into D-gram gold, and to use a third magic to turn e gram gold into F-gram sand. Ron and Hermione, two of Harry's best friends, were delighted, but Ron and Hermione had a disagreement, and Ron believed that Harry could gain unlimited gold with these three kinds of magic, Hermione that he could not get infinitely much. Harry wants to know exactly who's right.

Input

Enter a 6 integer a,b,c,d,e,f
(0<=a,b,c,d,e,f<=1000)

Output

The

output of who is right, assuming Ron is on the output "Ron". Otherwise output "Hermione"

Example input

100 200 250 150 200 250100 50 50 200 200 1001 1 0 1 1 1100 1 100 1 0 1

Example output

Ronhermioneronron

Uploaded by

Acm_ Wang Ying

Idea: Forward gold "reverse gold" (Ron)

Forward: K=b/c*d

Reverse: q= (k/e) *f

w= (q/a) *b

p= (w/c) *d//Reverse Gold

Be aware that the input can have 0 so

1.a=0 BCD time Infinite gold Ron C=0&&d infinitely change out of gold Ron

2. When (a=0| | c=0| | e=0) &&b, D, F are also infinite gold

3. When (a==0&&b==0) | | (e==0&&f==0) | | (c==0&&d==0) There is a link in the middle of herminoe that cannot be converted

(The SYY code is simpler than I wrote)

#include <stdio.h>int main () {    double a,b,c,d,e,f;    while (~SCANF ("%lf%lf%lf%lf%lf%lf", &a,&b,&c,&d,&e,&f))    {        if (c==0&&d)        {            printf ("ron\n");            Continue;        }        if (a==0&&b&&c&&d)        {            printf ("ron\n");            Continue;        }        if (a==0| | c==0| | e==0) (&&b&&d&&f)        {            printf ("ron\n");            Continue;        }        if ((a==0&&b==0) | | | (e==0&&f==0) | | (c==0&&d==0))        {            printf ("hermione\n");            Continue;        }        Double k= (b/c) *d;        Double q= (k/e) *f;        Double w= (q/a) *b;        Double p= (w/c) *d;        if (p>k)            printf ("ron\n");        else            printf ("hermione\n");    }    return 0;}  

Nyoj 949 Harry Potter (detail question)