USACO2.1.2--Ordered Fractions

Ordered Fractions

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

题解：枚举分子和分母。分子的范围为[1,n),分母的范围为(1,n]。假设两个分数a/b和c/d，如果gcd(a,b)==1那么就是符合题目要求的。枚举出符合要求的分数之后需要进行排序，判断a/b是否大于c/d，只需判断a*d>b*c是否成立即可。

 

View Code

 1 /*

 2 ID:spcjv51

 3 PROG:frac1

 4 LANG:C

 5 */

 6 #include<stdio.h>

 7 #include<math.h>

 8 #include<stdlib.h>

 9 int a[30000],b[30000];

10 int gcd(int x,int y)

11 {

12     if(x==0) return 0;

13     if(x%y==0) return y;

14     else

15     return(gcd(y,x%y));

16 }

17 int main(void)

18 {

19     freopen("frac1.in","r",stdin);

20     freopen("frac1.out","w",stdout);

21     int i,j,ans,n,temp;

22     scanf("%d",&n);

23     ans=0;

24     for(i=1;i<n;i++)

25         for(j=i+1;j<=n;j++)

26             if(gcd(i,j)==1)

27             {

28                 ans++;

29                 a[ans]=i;

30                 b[ans]=j;

31             }

32     for(i=1;i<ans;i++)

33     for(j=i+1;j<=ans;j++)

34     if(a[i]*b[j]>a[j]*b[i])

35     {

36         temp=a[i];

37         a[i]=a[j];

38         a[j]=temp;

39         temp=b[i];

40         b[i]=b[j];

41         b[j]=temp;

42     }

43     printf("0/1\n");

44     for(i=1;i<=ans;i++)

45     printf("%d/%d\n",a[i],b[i]);

46     printf("1/1\n");

47     return 0;

48 }

 

官网的题解，效率好高！！！

我们可以把0/1和1/1作为“端点”，通过把两个分数的分子相加、分母相加得到的新分数作为中点来递归（如图）

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

每一个分数的分子和分母都是由求和得来的，这意味着我们可以通过判断和与N的大小关系来判断递归边界。

<这是分数加成法！？数学上用于找一个无理数的近似分数>

#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");

}