USACO Section 2.1 Ordered Fractions（枚举）

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

分析：枚举、排序
心得：排序的时候起初用的是除，最后搞得出现精度问题，切记能不除就尽量不除。

/*
ID: dizzy_l1
LANG: C++
TASK: frac1
*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define MAXN 40000

using namespace std;

struct ANS
{
    int a,b;
}ans[MAXN];

bool cmp(ANS A,ANS B)
{
    if(A.a*B.b<A.b*B.a) return true;
    if(A.a*B.b==A.b*B.a&&A.a<B.a) return true;
    return false;
}

int main()
{
//    freopen("frac1.in","r",stdin);
//    freopen("frac1.out","w",stdout);
    int n,i,j,k;
    while(scanf("%d",&n)==1)
    {
        k=0;
        ans[k].a=0;ans[k].b=1;
        k++;
        for(i=1;i<=n;i++)
        {
            for(j=i;j<=n;j++)
            {
                ans[k].a=i;ans[k].b=j;
                k++;
            }
        }
        sort(ans,ans+k,cmp);
        printf("%d/%d\n",ans[0].a,ans[0].b);
        for(i=1;i<k;i++)
        {
            if(ans[i].a*ans[i-1].b==ans[i].b*ans[i-1].a)
                continue;
            printf("%d/%d\n",ans[i].a,ans[i].b);
        }
    }
    return 0;
}