USACO2.1.2Ordered 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。

/*
ID:spcjv51
PROG:frac1
LANG:C
*/
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int a[30000],b[30000];
int gcd(int x,int y)
{
    if(x==0) return 0;
    if(x%y==0) return y;
    else
    return(gcd(y,x%y));
}
int main(void)
{
    freopen("frac1.in","r",stdin);
    freopen("frac1.out","w",stdout);
    int i,j,ans,n,temp;
    scanf("%d",&n);
    ans=0;
    for(i=1;i<n;i++)
        for(j=i+1;j<=n;j++)
            if(gcd(i,j)==1)
            {
                ans++;
                a[ans]=i;
                b[ans]=j;
            }
    for(i=1;i<ans;i++)
    for(j=i+1;j<=ans;j++)
    if(a[i]*b[j]>a[j]*b[i])
    {
        temp=a[i];
        a[i]=a[j];
        a[j]=temp;
        temp=b[i];
        b[i]=b[j];
        b[j]=temp;
    }
    printf("0/1\n");
    for(i=1;i<=ans;i++)
    printf("%d/%d\n",a[i],b[i]);
    printf("1/1\n");
    return 0;
}

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");
}