USACO 2.1 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

题目大意：就是说给定一个N，输出值在0到1之间的，分母在1到N之间的所有值不重复的分数（可以约分的需要约分）。

思路：很简单，因为数据量小，所以就是枚举分子分母，然后不要不是最简分数的分数，再排序。

 /*
 ID:fffgrdc1
 PROB:frac1
 LANG:C++
 */
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 int prime[],primecnt=;
 bool bo[]={};
 struct str
 {
     int x;int y;
     double ans;
 }e[];
 bool kong(str aa,str bb)
 {
     return aa.ans<bb.ans;
 }
 bool check(int x,int y)
 {
     //int temp=sqrt(double (y));
     for(int i=;i<=primecnt&&prime[i]<=y;i++)
     {
         if(!(y%prime[i]))
             if(!(x%prime[i]))
                 return ;
     }
     return ;
 }
 int main()
 {
     freopen("frac1.in","r",stdin);
     freopen("frac1.out","w",stdout);
     int n;
     scanf("%d",&n);
     int anscnt=;
     bo[]=bo[]=;
     for(int i=;i<;i++)
     {
         if(!bo[i])
         {
             prime[++primecnt]=i;
             for(int j=;j*i<;j++)
             {
                 bo[i*j]=;
             }
         }
     }
     for(int i=;i<=n;i++)
     {
         for(int j=;j<i;j++)
         {
             if(check(j,i))
             {
                 e[++anscnt].x=j;
                 e[anscnt].y=i;
                 e[anscnt].ans=(j*1.0)/i;
             }
         }
     }
     sort(e+,e++anscnt,kong);
     printf("0/1\n");
     for(int i=;i<=anscnt;i++)
     {
         printf("%d/%d\n",e[i].x,e[i].y);
     }
     printf("1/1\n");
     return ;
 }

check函数写的太烂了。。。WA了几发都是因为想优化它。本来是想到用GCD的，但是担心时间复杂度的问题，后来学长告诉我不用担心呀，而且甚至不用自己手写，algorithm里面有现成的。。。于是代码变成下面这样也A了，而且复杂度下降了。。。。惊了

 /*
 ID:fffgrdc1
 PROB:frac1
 LANG:C++
 */
 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 int prime[],primecnt=;
 bool bo[]={};
 struct str
 {
     int x;int y;
     double ans;
 }e[];
 bool kong(str aa,str bb)
 {
     return aa.ans<bb.ans;
 }
 bool check(int x,int y)
 {
     return __gcd(x,y)==;
 }
 int main()
 {
     freopen("frac1.in","r",stdin);
     freopen("frac1.out","w",stdout);
     int n;
     scanf("%d",&n);
     int anscnt=;
     bo[]=bo[]=;
     for(int i=;i<;i++)
     {
         if(!bo[i])
         {
             prime[++primecnt]=i;
             for(int j=;j*i<;j++)
             {
                 bo[i*j]=;
             }
         }
     }
     for(int i=;i<=n;i++)
     {
         for(int j=;j<i;j++)
         {
             if(check(j,i))
             {
                 e[++anscnt].x=j;
                 e[anscnt].y=i;
                 e[anscnt].ans=(j*1.0)/i;
             }
         }
     }
     sort(e+,e++anscnt,kong);
     printf("0/1\n");
     for(int i=;i<=anscnt;i++)
     {
         printf("%d/%d\n",e[i].x,e[i].y);
     }
     printf("1/1\n");
     return ;
 }