this is the running code to find no of triangles using brute force technique
logic: in a triangle having sides a,b,c; then a+b>c(if c is greatest side)
correct me if i am wrong.
#include<iostream>
#include<conio.h>
using namespace std;
int max(int a,int b,int c)
{
return ((a>b?a:b)>c?(a>b?a:b):c);
}
int main()
{
int n,a[120],t,p,q;
cin>>t;
while(t--)
{
cin>>n;
for(int i=0;i<n;i++)
cin>>a[i];
int i,j,no_of_triangles=0,sum=0,largest_edge;
for(i=0;i<n-2;i++)
{
for(j=i+2;j<n;j++)
{
sum=a[i]+a[i+1]+a[j];
largest_edge=max(a[i],a[i+1],a[j]);
if(sum-largest_edge>largest_edge)
{
no_of_triangles++;
cout<<a[i]<<" "<<a[i+1]<<" "<<a[j]<<"\n";}
}
}
cout<<no_of_triangles<<endl;
}
getch();
return 0;
}
On Wednesday, 13 June 2012 22:18:01 UTC+5:30, payel roy wrote:
>
>
> Let's say there are N sides are given. Length of them are like
> 1,2,3,4,5,....N.
>
> How do you determine how many tri-angles can be made out of these N sides?
>
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