Geometry Forever


This is a companion discussion topic for the original entry at https://toph.co/p/geometry-forever

#include <stdio.h>

int main() {
int A,B,sum;
scanf("%d %d",&A,&B);
sum=A+B;
printf("%d\n",sum-1);

return 0;

}

what’s wrong with this code?
#always correct in 1st case but wrong after 2nd case!

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Not of the first two.
If the give lengths of two side are a and b and the third side is c then a + b > c , a + c > b and b + c > a. Your code will work for the first one but not for the others.

#include<stdio.h>

int main()
{
int A, B, count=0, C;
scanf("%d %d", &A, &B);
if(A>=1 && B<=100)
{
for(C=A-B; C<A+B; C++)
{
if(C>A-B && C<A+B)
{
count++;
}
}
printf("%d\n", count);
}
return 0;
}

now what’s the problem in this code?

I think this is a fault in the problem. As far as it says, there should be the values sides A and B given and we find the possible number of integer values of side C. But from your argument, it seems like the values of A and B are given and answers are checked by the values of B and C or whatsoever. Correct me if I’m wrong.

Actually I don’t understand what are you saying but it seems you didn’t understand what I meant. Suppose that, A = 5, B = 3. Accordingly to backbenchermmc’s code The answer should be 5+3-1 = 7 and the set of the answers should be {1, 2, 3, 4, 5, 6, 7}.
A+B=5+3=8 and it is greater than any of the value of the set
Now, this answer satisfies A+B>C
But there are two more conditions. A+C>B and B+C>A
Now, for the first value B+C = 3+1 = 4
Which isn’t greater then 5
So the actual set of answer is {3, 4, 5, 6, 7}
And the actual answer is 5

#include <bits/stdc++.h>
#define ll long long
#define ff first
#define ss second
using namespace std;

ll def(ll a, ll b) {
if(a>b) return a-b;
return b-a;
}

int main() {
ll a , b;
cin >> a >> b;
cout << (a+b) - (def(a,b)+1);
}

//Maybe it is the shortest method

The shortest solution that I provide is here:::

>     #include <bits/stdc++.h>
>     using namespace std;
> 
>     int main()
>     {
>         int a, b;
>         cin >> a >> b;
>         int minimum = min(a, b);
>         cout << (2 * minimum - 1) << endl;
>     }

Thank you all!

another possible answer is here:

#include
using namespace std;
int main()
{
int i,j,a,b;
cin>>a>>b;
i=min(a,b);
j=max(a,b);
cout<<(a+b)-((j-i)+1)<<endl;
return 0;
}

What is the problem?

#include<stdio.h>

int main()
{
int N,X;
int c=0;
scanf("%d %d",&N,&X);

for(int i=1;i<(N+X);i++){
	if (N<(i+X) && X<(i+N)){
	c++
	
	}
	
}
	printf("%d\n ",c);

return 0;

}

Hi Guys,

If two sides of a triangle are given the probale third side should
statify the following condition:

 - Side A + Side B > Side C 
 - Side A + Side C > Side B 
 - Side C + Side B > Side A
 - Area of the triangle > 0
    Area of the triangle = Math.sqrt(s * (s-Side A) * (s-Side B) * (s-Side C));
    s-value = (Side A + Side B + Side C)/2
    Reference: https://www.mathsisfun.com/geometry/herons-formula.html
   
FOR INSTANT
-----------

Side A = 3                                                                                                                               
Side B = 5                                                                                                                               

Side C	s-Value	Area                                                                                                             
  7 	 7.5 	6.49519052838329                                                                                                          
  6 	 7.0 	7.483314773547883                                                                                                         
  5 	 6.5 	7.1545440106270926                                                                                                        
  4 	 6.0 	6.0                                                                                                                       
  3 	 5.5 	4.14578098794425                                                                                                          

PROBABLE Side C = 3 4 5 6 7        


IN/OUTPUT SET
-------------
1 1	1
3 5	3 4 5 6 7
12 15	4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
56 66	11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 
	32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 
	53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 
	74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 
	95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 
	112 113 114 115 116 117 118 119 120 121

There could be multiple values of Side C in most cases. And, these sets of 
outcome satify all pre-condition required to find Side C. The output sets need
to be corrected!
	
SAMPLE CODE IN JAVA
--------------------
import java.util.Scanner;
import java.lang.Math; 

public class Solution {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int sideA = sc.nextInt();
		int sideB = sc.nextInt();
		int sumAB = sideA + sideB;
		double s=0F, area=0F;
		int arr[] = new int[sumAB];
		int count = 0;
		if(0<sideA && sideA<= 100 && 0<sideB && sideB<=100){
		    for(int i=sumAB; i>=1; i--){
		        s = (sumAB+i)/2F;
		        area = Math.sqrt(s * (s-sideA) * (s-sideB) * (s-i));
		        arr[count] = 0;
		        if(area > 0F){
		            arr[count] = i;
		        }
		        count++;
		    }
		    for(int j=count-1; j>=0; j--){
		        if(arr[j]>0){
		            if(((sideA + sideB) > arr[j]) && ((sideA + arr[j]) > sideB) && ((arr[j] + sideB) > sideA)){
		                System.out.println(arr[j]);
		            }
		        }
		    }
		}
		//The output sets are incorrect!!!
	}
}

I hope things are understood and the output sets will be corrected.
Thanks.