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This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

Consider the set of all reduced fractions between 0 and 1 inclusive with denominators less than or equal to N.

Here is the set when N = 5:

0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1

Write a program that, given an integer N between 1 and 160 inclusive, prints the fractions in order of increasing magnitude.

PROGRAM NAME: frac1

INPUT FORMAT

One line with a single integer N.

SAMPLE INPUT (file frac1.in)

5

OUTPUT FORMAT

One fraction per line, sorted in order of magnitude.

SAMPLE OUTPUT (file frac1.out)

0/1
1/5
1/4
1/3
2/5
1/2
3/5
2/3
3/4
4/5
1/1

CODE

Java

C++

Pascal

ANALYSIS

Alex Schwedner

Here’s a very fast, straightforward solution from Alex Schwedner:

#include <fstream.h>
#include <stdlib.h>

struct fraction {
    int numerator;
    int denominator;
};

bool rprime(int a, int b){
   int r = a % b;
   while(r != 0){
       a = b;
       b = r;
       r = a % b;
   }
   return(b == 1);
}

int fraccompare (struct fraction *p, struct fraction *q) {
   return p->numerator * q->denominator - p->denominator *q->numerator;
}

int main(){
   int found = 0;
   struct fraction fract[25600];

   ifstream filein("frac1.in");
   int n;
   filein >> n;
   filein.close();

   for(int bot = 1; bot <= n; ++bot){
       for(int top = 0; top <= bot; ++top){
           if(rprime(top,bot)){
               fract[found].numerator = top;
               fract[found++].denominator = bot;
           }
       }
   }

   qsort(fract, found, sizeof (struct fraction), fraccompare);

   ofstream fileout("frac1.out");
   for(int i = 0; i < found; ++i)
       fileout << fract[i].numerator << '/' << fract[i].denominator << endl;
   fileout.close();

   exit (0);
}

Russ Cox

Here’s a super fast solution from Russ:

We notice that we can start with 0/1 and 1/1 as our ‘endpoints’ and recursively generate the middle points by adding numerators and denominators.

0/1                                                              1/1
                               1/2
                  1/3                      2/3
        1/4              2/5         3/5                 3/4
    1/5      2/7     3/8    3/7   4/7   5/8       5/7         4/5

Each fraction is created from the one up to its right and the one up to its left. This idea lends itself easily to a recursion that we cut off when we go too deep.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

int n;
FILE *fout;

/* print the fractions of denominator <= n between n1/d1 and n2/d2 */
void
genfrac(int n1, int d1, int n2, int d2)
{
	if(d1+d2 > n)	/* cut off recursion */
		return;

	genfrac(n1,d1, n1+n2,d1+d2);
	fprintf(fout, "%d/%d\n", n1+n2, d1+d2);
	genfrac(n1+n2,d1+d2, n2,d2);
}

void
main(void)
{
	FILE *fin;

	fin = fopen("frac1.in", "r");
	fout = fopen("frac1.out", "w");
	assert(fin != NULL && fout != NULL);

	fscanf(fin, "%d", &n);

	fprintf(fout, "0/1\n");
	genfrac(0,1, 1,1);
	fprintf(fout, "1/1\n");
}

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