USACO 2.1 Ordered Fractions

很简单的一道题

直接找出所有的分数，然后按值排序，去除重复的输出。

/*
ID: superbi1
LANG: C
TASK: frac1
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define NL 160*161
#define EP 1e-10

struct Fct{
    int a, b;
    double val;
}f[NL];
int nf;

int gcd(int a, int b)
{
    if (b == 0) return a;
    return gcd(b, a%b);
}

int cmp(const void *a, const void *b)
{
    struct Fct *x = (struct Fct *)a;
    struct Fct *y = (struct Fct *)b;
    if (fabs(x->val - y->val) < EP) return 0;
    else if (x->val < y->val) return -1;
    else return 1;
}

int cmp1(struct Fct *a, struct Fct *b)
{
    return a->a - a->b;
}

int main()
{
    FILE *in = fopen("frac1.in", "r");
    FILE *out = fopen("frac1.out", "w");
    int n;
    int I, K;
    int a, b;
    fscanf(in, "%d", &n);
    nf = 0;
    f[nf].a = 0;
    f[nf].b = 1;
    f[nf++].val = 0;
    f[nf].a = 1;
    f[nf].b = 1;
    f[nf++].val = 1;
    for (I=1; I<=n; I++) {
        for (K=1; K<I; K++) {
            int g = gcd(I, K);
            f[nf].a = K/g;
            f[nf].b = I/g;
            f[nf++].val = K*1.0/I;
        }
    }
    qsort(f, nf, sizeof(f[0]), cmp);
    a = f[0].a;
    b = f[0].b;
    for (I=1; I<nf; I++) {
        if (a==f[I].a && b==f[I].b) continue;
        else {
            fprintf(out, "%d/%d\n", a, b);
            a = f[I].a;
            b = f[I].b;
        }
    }
    fprintf(out, "%d/%d\n", a, b);
    return 0;
}

标程一：

改进：

1.发现这些分数的分子和分母互素

2.排序的时候的小技巧：p1/q1 = p2/q2 --> p1*q2 = p2*q1

#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);
}

标程二：

改进：

1.找出这些分数之间的规律，然后直接生成

#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");
}