How Do I Make Counting In Nested Loops As Fast As Possible?

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Im inputing three numbers of users choice
number 1
number 2
final number

what i want to find out is , how many combinations of number 1 and number 2 would result in final number = number1 *x + number2*y = final number.

I am using nested loops

C

int i;
int j;
long long int pocet=0; // the numbers of combination
long long int one; // first number inputed by user
long long int two; // second number inputed by user
long long int last; // final number inputed by user - wanted result
long long int one_max;
long long int two_max;
long long int add=1; //for incrementing loops
long long int reduc=1; // for decrementing loops

if((one%2==0) && (two%2==0)){
        if(one>two){
            add=two/2;
            reduc=one/2;
    }
}
one_max=last/one;
two_max=last/two;
for(i=0;i<=one_max;i+=add){
    for(j=two_max;j>=0;j-=reduc){
        long long int tmp_one=(one*i);
        long long int tmp_two=(two*j);
            if(tmp_one+tmp_two==last){
                two_max=j;
                pocet++;
        }
    }
}

if i input these numbers
one = 10;
two = 13;
final =100;
the only possible combination of first two numbers to result in final number is
10 * 10 + 13 * 0
The numbers have to be non negative

Posted 7-Nov-15 4:56am

Member 11378302

Updated 7-Nov-15 6:24am

v5

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Comments

PIEBALDconsult 7-Nov-15 11:12am

That's hard to read with it's inconsistent indenting odd words thrown in. It doesn't look like it would even compile.
Please use Improve question to format the code consistently and add more detail about what you are trying to do and why.

Member 11378302 7-Nov-15 11:25am

i have updated question.

nv3 7-Nov-15 12:04pm

Your statement is not true:

10 * -120 + 13 * 100 = 100

So there is obviously more than one combination of x any that fits the rule. Or you should have stated that x and y need to be non-negative.

Member 11378302 7-Nov-15 12:25pm

indeed , i have updated the question for this info.

BillWoodruff 7-Nov-15 16:17pm

Fix your code so it compiles.

2 solutions

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Richard MacCutchan · Answer 1 · 2015-11-07T05:10:00

Solution 1

You already posted this question in the Algorithms forum. Please post in one place only.

Posted 7-Nov-15 5:10am

Richard MacCutchan

Patrice T · Answer 2 · 2015-11-07T08:14:00

Here is code with first batch of changes. the code is changed to a function for practical reasons.

C++

long long int MyFunc (long long int one, long long int two, long long int last) {
	// one; // first number inputed by user
	// two; // second number inputed by user
	// int last; // final number inputed by user - wanted result
	
	// Question: How many (i,j) pairs (as positive integers) will verify
	// one*i+ two*j = last (constant part is on right)
	
	long long int i;
	long long int j;
	long long int pocet=0; // the numbers of combination
	
	for(i=0;i*one <=last;i++){
		// two*j = last - one*i (constant part is on right)
		// j = (last - one*i) / two (constant part is on right)
		if ((last - one*i )% two == 0) { // integer ?
			// solution
			j= (last - one*i) / two;
			pocet++;
		}
	}
	return pocet;
}

I just stick to the problem here.
I just take advantage of mathematics to change the equation as i is known. This should already improve the runtime.