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。

 1 /*
 2 ID:fffgrdc1
 3 PROB:frac1
 4 LANG:C++
 5 */
 6 #include<cstdio>
 7 #include<iostream>
 8 #include<algorithm>
 9 #include<cmath>
10 using namespace std;
11 int prime[100],primecnt=0;
12 bool bo[200]={0};
13 struct str
14 {
15     int x;int y;
16     double ans;
17 }e[40000];
18 bool kong(str aa,str bb)
19 {
20     return aa.ans<bb.ans;
21 }
22 bool check(int x,int y)
23 {
24     //int temp=sqrt(double (y));
25     for(int i=1;i<=primecnt&&prime[i]<=y;i++)
26     {
27         if(!(y%prime[i]))
28             if(!(x%prime[i]))
29                 return 0;
30     }
31     return 1;
32 }
33 int main()
34 {
35     freopen("frac1.in","r",stdin);
36     freopen("frac1.out","w",stdout);
37     int n;
38     scanf("%d",&n);
39     int anscnt=0;
40     bo[0]=bo[1]=1;
41     for(int i=2;i<200;i++)
42     {
43         if(!bo[i])
44         {
45             prime[++primecnt]=i;
46             for(int j=2;j*i<200;j++)
47             {
48                 bo[i*j]=1;
49             }
50         }
51     }
52     for(int i=1;i<=n;i++)
53     {
54         for(int j=1;j<i;j++)
55         {
56             if(check(j,i))
57             {
58                 e[++anscnt].x=j;
59                 e[anscnt].y=i;
60                 e[anscnt].ans=(j*1.0)/i;
61             }
62         }
63     }
64     sort(e+1,e+1+anscnt,kong);
65     printf("0/1\n");
66     for(int i=1;i<=anscnt;i++)
67     {
68         printf("%d/%d\n",e[i].x,e[i].y);
69     }
70     printf("1/1\n");
71     return 0;
72 }

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

 1 /*
 2 ID:fffgrdc1
 3 PROB:frac1
 4 LANG:C++
 5 */
 6 #include<cstdio>
 7 #include<iostream>
 8 #include<algorithm>
 9 #include<cmath>
10 using namespace std;
11 int prime[100],primecnt=0;
12 bool bo[200]={0};
13 struct str
14 {
15     int x;int y;
16     double ans;
17 }e[40000];
18 bool kong(str aa,str bb)
19 {
20     return aa.ans<bb.ans;
21 }
22 bool check(int x,int y)
23 {
24     return __gcd(x,y)==1;
25 }
26 int main()
27 {
28     freopen("frac1.in","r",stdin);
29     freopen("frac1.out","w",stdout);
30     int n;
31     scanf("%d",&n);
32     int anscnt=0;
33     bo[0]=bo[1]=1;
34     for(int i=2;i<200;i++)
35     {
36         if(!bo[i])
37         {
38             prime[++primecnt]=i;
39             for(int j=2;j*i<200;j++)
40             {
41                 bo[i*j]=1;
42             }
43         }
44     }
45     for(int i=1;i<=n;i++)
46     {
47         for(int j=1;j<i;j++)
48         {
49             if(check(j,i))
50             {
51                 e[++anscnt].x=j;
52                 e[anscnt].y=i;
53                 e[anscnt].ans=(j*1.0)/i;
54             }
55         }
56     }
57     sort(e+1,e+1+anscnt,kong);
58     printf("0/1\n");
59     for(int i=1;i<=anscnt;i++)
60     {
61         printf("%d/%d\n",e[i].x,e[i].y);
62     }
63     printf("1/1\n");
64     return 0;
65 }