USACO Section 2.1 Ordered Fractions 解题报告

题目

题目描述

给定一个数N(1<=N<=160)，需要产生所有的分数，这些分数的值必须要在0~1之间。而且每个分数的分母不能超过N。如下例所示：

N = 5
产生所有的分数：0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1

样例输入

5

样例输出

0/1
1/5
1/4
1/3
2/5
1/2
3/5
2/3
3/4
4/5
1/1

解题思路

这个题目最一开始我走了一些弯路，想得太复杂，结果第一发就超时了。后来静下心来算了想了下，发现枚举加排序速度就已经很快了。所以就实现了一下，通过。

解题代码

/*
ID: yinzong2
PROG: frac1
LANG: C++11
*/
#define MARK
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 170;
int N;
struct Frac {
    int num;
    int deno;
};
vector<Frac> lst;

bool cmp(Frac A, Frac B) {
    int temp1 = A.num*B.deno;
    int temp2 = A.deno*B.num;
    return temp1 < temp2;
}

int GCD(int x, int y) {
    if (0 == y) return x;
    return GCD(y, x%y);
}

Frac afterGCD(Frac f) {
    int gcd = GCD(f.num, f.deno);
    f.num /= gcd;
    f.deno /= gcd;
    return f;
}

int main() {
#ifdef MARK
    freopen("frac1.in", "r", stdin);
    freopen("frac1.out", "w", stdout);
#endif // MARK
    while (cin >> N) {
        cout << "0/1" << endl;
        if (N != 1) {
            lst.clear();
            for (int i = N; i >= 2; --i) {
                for (int j = 1; j < i; ++j) {
                    Frac f;
                    f.num = j;
                    f.deno = i;
                    lst.push_back(f);
                }
            }
            sort(lst.begin(), lst.end(), cmp);
            int len = lst.size();
            Frac pre;
            pre = afterGCD(lst[0]);
            cout << pre.num << "/" << pre.deno << endl;
            // 对于排序后的数据进行去重
            for (int i = 1; i < len; ++i) {
                lst[i] = afterGCD(lst[i]);
                if (lst[i].num == pre.num && lst[i].deno == pre.deno) continue;
                cout << lst[i].num << "/" << lst[i].deno << endl;
                pre = lst[i];
            }
        }
        cout << "1/1" << endl;
    }
}
/*
Executing...
   Test 1: TEST OK [0.000 secs, 4180 KB]
   Test 2: TEST OK [0.000 secs, 4180 KB]
   Test 3: TEST OK [0.000 secs, 4180 KB]
   Test 4: TEST OK [0.014 secs, 4180 KB]
   Test 5: TEST OK [0.014 secs, 4180 KB]
   Test 6: TEST OK [0.014 secs, 4180 KB]
   Test 7: TEST OK [0.028 secs, 4180 KB]
   Test 8: TEST OK [0.070 secs, 4180 KB]
   Test 9: TEST OK [0.056 secs, 4180 KB]
   Test 10: TEST OK [0.056 secs, 4188 KB]
   Test 11: TEST OK [0.140 secs, 4188 KB]

All tests OK.
*/

解题思路（type2）

之后看了看官方的题解，发现第一个解法就是用的枚举加排序，但是有个优化的地方我之前没有考虑。我们最终枚举出来的所有的分数，分子分母应该都是互质的。所以我们对于所有分子分母互质的分数进行排序输出即可。

解题代码

/*
ID: yinzong2
PROG: frac1
LANG: C++11
*/
#define MARK
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 170;
int N;
struct Frac {
    int num;
    int deno;
};
vector<Frac> lst;

bool cmp(Frac A, Frac B) {
    int temp1 = A.num*B.deno;
    int temp2 = A.deno*B.num;
    return temp1 < temp2;
}

int GCD(int x, int y) {
    if (0 == y) return x;
    return GCD(y, x%y);
}

int main() {
#ifdef MARK
    freopen("frac1.in", "r", stdin);
    freopen("frac1.out", "w", stdout);
#endif // MARK
    while (cin >> N) {
        cout << "0/1" << endl;
        if (N != 1) {
            lst.clear();
            for (int i = N; i >= 2; --i) {
                for (int j = 1; j < i; ++j) {
                    Frac f;
                    f.num = j;
                    f.deno = i;
                    if (GCD(f.num, f.deno) != 1) continue;
                    lst.push_back(f);
                }
            }
            sort(lst.begin(), lst.end(), cmp);
            int len = lst.size();
            for (int i = 0; i < len; ++i) {
                cout << lst[i].num << "/" << lst[i].deno << endl;
            }
        }
        cout << "1/1" << endl;
    }
}
/*
Executing...
   Test 1: TEST OK [0.000 secs, 4180 KB]
   Test 2: TEST OK [0.000 secs, 4180 KB]
   Test 3: TEST OK [0.000 secs, 4180 KB]
   Test 4: TEST OK [0.000 secs, 4180 KB]
   Test 5: TEST OK [0.000 secs, 4180 KB]
   Test 6: TEST OK [0.000 secs, 4180 KB]
   Test 7: TEST OK [0.000 secs, 4180 KB]
   Test 8: TEST OK [0.014 secs, 4180 KB]
   Test 9: TEST OK [0.028 secs, 4180 KB]
   Test 10: TEST OK [0.056 secs, 4180 KB]
   Test 11: TEST OK [0.140 secs, 4188 KB]

All tests OK.
*/

解题思路（type3）

官方第二个解法很巧妙。是找到一个数学规律，直接可以打印所有的数字，时间复杂度为O(N)。具体可以看这里。

解题代码

/*
ID: yinzong2
PROG: frac1
LANG: C++11
*/
#define MARK
#include <iostream>
#include <cstdio>
using namespace std;

int N;

void generateFrac(int num1, int denom1, int num2, int denom2) {
    if (denom1 + denom2 > N) return ;
    generateFrac(num1, denom1, num1+num2, denom1+denom2);
    cout << num1+num2 << "/" << denom1+denom2 << endl;
    generateFrac(num1+num2, denom1+denom2, num2, denom2);
}

int main() {
#ifdef MARK
    freopen("frac1.in", "r", stdin);
    freopen("frac1.out", "w", stdout);
#endif // MARK
    while (cin >> N) {
        cout << "0/1" << endl;
        generateFrac(0, 1, 1, 1);
        cout << "1/1" << endl;
    }
    return 0;
}
/*
Executing...
   Test 1: TEST OK [0.000 secs, 4176 KB]
   Test 2: TEST OK [0.000 secs, 4176 KB]
   Test 3: TEST OK [0.000 secs, 4176 KB]
   Test 4: TEST OK [0.000 secs, 4176 KB]
   Test 5: TEST OK [0.000 secs, 4176 KB]
   Test 6: TEST OK [0.000 secs, 4176 KB]
   Test 7: TEST OK [0.000 secs, 4176 KB]
   Test 8: TEST OK [0.014 secs, 4176 KB]
   Test 9: TEST OK [0.028 secs, 4176 KB]
   Test 10: TEST OK [0.056 secs, 4176 KB]
   Test 11: TEST OK [0.140 secs, 4176 KB]

All tests OK.
*/