CODEWITHSUNDEEP
c++
Culver Kwan
Culver Kwan

Reputation: 111

Debugging code for linear equations with two unknowns

On a judge platform, I encountered this problem, which I had wrong answer in 3 out of the 18 tests:

In the simultaneous equations

ax+by=c

dc+ey=f

x and y are unknowns. The coefficients a, b, c, d, e and f are integers. Each of the coefficients ranges from -2000 to 2000.

Write a program to read in the coefficients and solve for the unknowns of the corresponding simultaneous equations.

INPUT

The first and the only line contains six integers a, b, c, d, e and f separated by spaces.

OUTPUT

If there is exactly one solution, the first and the only line contains two numbers x and y, separated by a space, so that when they are substituted to the original equations, the set of simultaneous equations hold. If the set of simultaneous equations has no solution, display no solution. If the set of simultaneous equations has more than one solution, display many solutions. Assume that x and y are always integers.

So, this is my code:

#include<bits/stdc++.h>
using namespace std;
int det(int a,int b,int c,int d){
    return a*d-b*c;
}
int main(){
    int a,b,c,d,e,f,det1,detx,dety;
    cin >> a >> b >> c >> d >> e >> f;
    det1=det(a,b,d,e);
    detx=det(c,b,f,e);
    dety=det(a,c,d,f);
    if(det1==0){
        if(detx==0){
            cout << "many solutions";
        }
        else{
            cout << "no solution";
        }
    }
    else{
        cout << detx/det1 << ' ' << dety/det1;
    }
}

Can someone help me to debug this code?

Upvotes: 1

Views: 141

Answers (1)

A M
A M

Reputation: 15265

My big guess is that I do not understand the question. A linear equation system with 2 unknown looks normally different.

If this is really the question, then we can do the following:

ax+by=c
dc+ey=f

ey = f-dc
y=(f-dc)/e

ax = c-by
   = c-b((f-dc)/e)
   = c-b(f-dc)/e
   = c-(bf+bcd)/e
   = c-bf/e+bcd/e
 
x = c/a-bf/(ae)+bcd/(ae)
  = c/a-bf/a/e+bcd/a/e

So except with "e" or "a" being 0 there is always one deterministic solution.

That is so trivial, that I am nearly persuaded that I do not understand the question.

Even with a linear dependency of a<->dc b<->e nothing would change. Mabye

ax+by=c
dx+ey=f

was intended?

Upvotes: 2

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