usaco-orderd fractions

//usaco training-Ordered Fractions/* * 题目类型：枚举 * 题意：输入一个自然数N，写一个程序来增序输出分母小于等于N的既约真分数 * 用双重循环搜索所有不同分子和分母的情况，将满足条件的分数保存起来 * 按分数的大小排序输出 *//*Compiling...Compile: OKExecuting... Test 1: TEST OK [0.000 secs, 3188 KB] Test 2: TEST OK [0.000 secs, 3188 KB] Test 3: TEST OK [0.000 secs, 3188 KB] Test 4: TEST OK [0.000 secs, 3188 KB] Test 5: TEST OK [0.000 secs, 3188 KB] Test 6: TEST OK [0.000 secs, 3188 KB] Test 7: TEST OK [0.000 secs, 3188 KB] Test 8: TEST OK [0.027 secs, 3188 KB] Test 9: TEST OK [0.081 secs, 3188 KB] Test 10: TEST OK [0.162 secs, 3188 KB] Test 11: TEST OK [0.432 secs, 3188 KB]All tests OK. */#include <iostream>#include <algorithm>#include <fstream>using namespace std;#define MAX 10000struct point{int a,b;double d;}pos[MAX];int gcd(int a,int b);int cmp(point x,point y);int main(void){int i,j,n,k=0;ifstream fin("frac1.in");ofstream fout("frac1.out");fin>>n;for(i=2; i<=n; ++i)for(j=1; j<i; ++j)if(gcd(i,j)==1){pos[k].a=i;pos[k].b=j;pos[k++].d=j*1.0/i;}sort(pos,pos+k,cmp);fout<<"0/1"<<endl;for(i=0; i!=k; ++i)fout<<pos[i].b<<'/'<<pos[i].a<<endl;fout<<"1/1"<<endl;return 0;}int gcd(int a,int b){return (b==0) ? a : gcd(b,a%b);}int cmp(point x,point y){return x.d<y.d;}