USACO 2.1.2 Ordered Fractions

一.题目描述

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

二.题目分析

由于题目是出现在第二章节的填充算法中的，一开始我尝试向BFS，DFS方向想，事实是并没有用。。。简单来想，只是一个枚举加互质判断的过程，由于最后需要有序输出，当然要用到qsort了。智商有限，只能想到这种肤浅的做法，官方给出了很好的答案，再次也给出，大家可以脑补一下哈。

三.代码

1.民间版

#include <stdio.h>
#include <stdlib.h>
#define MAX 13000
typedef struct fra
{
int a,b;   //a代表分子，b代表分母
}Frac;
int gcd(int x,int y)
{
if(x==0)
return y;
if(y==0)
return x;
if(x<y)
return gcd(y,x);
return gcd(y,x%y);
}
int cmp(const void*a,const void*b)
{
Frac x1=*(Frac*)a;
Frac x2=*(Frac*)b;

int t1=x1.b,t2=x2.b;   //一定要先把原分母存起来，否则下面会乘乱的，第一次就是这么出错的。

x1.b=x1.b*t2;
x1.a=x1.a*t2;

x2.b=x2.b*t1;
x2.a=x2.a*t1;

return x1.a-x2.a;
}
int main()
{
int N,i,j,k;
Frac f[MAX];
FILE *in=fopen("frac1.in","r"),*out=fopen("frac1.out","w");

if(!in||!out)
{
printf("file open error!\n");
return -1;
}

fscanf(in,"%d",&N);
k=0;
for(i=2;i<=N;i++)
{
for(j=1;j<i;j++)
{
if(gcd(i,j)==1)
{
f[k].a=j;
f[k].b=i;
k++;
}
}
}

qsort(f,k,sizeof(f[0]),cmp);

fprintf(out,"0/1\n");
for(i=0;i<k;i++)
fprintf(out,"%d/%d\n",f[i].a,f[i].b);
fprintf(out,"1/1\n");

return 0;
}

2.官方

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");
}