usaco2.1Ordered Fractions( 枚举， 数学）

这道题是给出一个n 求出分子分母<=n 且 gcd(u,d) = 1 ,  (u代表分子，d代表分母） 0<=u/d<=1

常规方法就是枚举n平方个数字。。然后进行比较排序等。。看了解答后发现一种数学方法比较神奇。 代码也非常短，递归版本

大致思路是 0/0  -------------------------1/1

 　　　　　0/0  ----------1/2----------1/1

　　　　　　0/0---1/4---1/2----2/3----1/1

如此这样递归下去。。具体数学证明我就不搬运了。。

题目：

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


代码：

/*
ID:doubles3
PROB:frac1
LANG:C++
*/
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<fstream>
#include<math.h>
#include <algorithm>
#include<string.h>
#include<string>
#include<vector>
#include <queue>
#include <set>
using namespace std;

int N;

int print(int au,int ad,int bu,int bd)
{
    if(ad+bd<=N)
    {
        print(au,ad,au+bu,ad+bd);
        cout<<au+bu<<"/"<<ad+bd<<endl;
        print(au+bu,ad+bd,bu,bd);
    }
    else
        return 0;
}

int main()
{
    freopen("frac1.in","r",stdin);
    freopen("frac1.out","w",stdout);

    cin>>N;

    cout<<"0/1"<<endl;
    print(0,1,1,1);
    cout<<"1/1"<<endl;

    return 0;
}