Usaco-section 2.1 Ordered Fractions (sort)

Description

Enter a natural number N, for a minimal fraction of A/b (numerator and denominator coprime fractions), to meet 1<=b<=n,0<=a/b<=1, find all the scores that meet the criteria.

Here is an example, when n=5, all solutions are:

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

Given a natural number n,1<=n<=160, programming outputs all the solutions in the order in which the fractional values increment.

Note: The greatest common divisor of ①0 and any natural number is the natural number ② coprime refers to the two natural numbers that greatest common divisor equals 1.

format

Program NAME: Frac1

INPUT FORMAT:

(File frac1.in)

Single line one natural number n (1..160)

OUTPUT FORMAT:

(File frac1.out)

Each score is a single row, in order of size

SAMPLE INPUT

5

SAMPLE OUTPUT

0/11/51/41/32/51/23/52/33/44/51/1

The data is small, and all the numbers are stored directly in the array, and then sorted.

Note: Only the simplest fraction is put into the array, because if it is not the simplest fraction, it will be able to be found in the array after simplification

/*id:your_id_hereprog:frac1lang:c++*/#include <cstdio> #include <algorithm>using namespace std;struct Fact {    int a, b;    BOOL operator < (const fact& x) const {        return a*x.b<x.a*b;    }} F[12999];int Main () {    int i,j,n,num;    Freopen ("Frac1.in", "R", stdin);    Freopen ("Frac1.out", "w", stdout);    while (1==SCANF ("%d", &n)) {        printf ("0/1\n");        for (Num=0,i=2;i<=n;++i) for            (j=1;j<i;++j) {                if (__GCD (i,j) ==1) {//did not look closely at the sample, in the Recruit ... If it is not the simplest fraction, it is not counted (the simplest fraction is already in the array)                    f[num].a=j;                    f[num++].b=i;                }            }        Sort (f,f+num);        for (i=0;i<num;++i)            printf ("%d/%d\n", f[i].a,f[i].b);        printf ("1/1\n");    }    return 0;}

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Usaco-section 2.1 Ordered Fractions (sort)