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[100],primecnt=0;
bool bo[200]={0};
struct str
{
    int x;int y;
    double ans;
}e[40000];
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=1;i<=primecnt&&prime[i]<=y;i++)
    {
        if(!(y%prime[i]))
            if(!(x%prime[i]))
                return 0;
    }
    return 1;
}
int main()
{
    freopen("frac1.in","r",stdin);
    freopen("frac1.out","w",stdout);
    int n;
    scanf("%d",&n);
    int anscnt=0;
    bo[0]=bo[1]=1;
    for(int i=2;i<200;i++)
    {
        if(!bo[i])
        {
            prime[++primecnt]=i;
            for(int j=2;j*i<200;j++)
            {
                bo[i*j]=1;
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<i;j++)
        {
            if(check(j,i))
            {
                e[++anscnt].x=j;
                e[anscnt].y=i;
                e[anscnt].ans=(j*1.0)/i;
            }
        }
    }
    sort(e+1,e+1+anscnt,kong);
    printf("0/1
");
    for(int i=1;i<=anscnt;i++)
    {
        printf("%d/%d
",e[i].x,e[i].y);
    }
    printf("1/1
");
    return 0;
}

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[100],primecnt=0;
bool bo[200]={0};
struct str
{
    int x;int y;
    double ans;
}e[40000];
bool kong(str aa,str bb)
{
    return aa.ans<bb.ans;
}
bool check(int x,int y)
{
    return __gcd(x,y)==1;
}
int main()
{
    freopen("frac1.in","r",stdin);
    freopen("frac1.out","w",stdout);
    int n;
    scanf("%d",&n);
    int anscnt=0;
    bo[0]=bo[1]=1;
    for(int i=2;i<200;i++)
    {
        if(!bo[i])
        {
            prime[++primecnt]=i;
            for(int j=2;j*i<200;j++)
            {
                bo[i*j]=1;
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<i;j++)
        {
            if(check(j,i))
            {
                e[++anscnt].x=j;
                e[anscnt].y=i;
                e[anscnt].ans=(j*1.0)/i;
            }
        }
    }
    sort(e+1,e+1+anscnt,kong);
    printf("0/1
");
    for(int i=1;i<=anscnt;i++)
    {
        printf("%d/%d
",e[i].x,e[i].y);
    }
    printf("1/1
");
    return 0;
}