USACO section 2.1 Ordered Fractions(简单数学求约数加个排序)

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
1/51/4
1/3
2/51/2
3/52/3
3/4
4/51/1

简单题：欧几里得求约数+快排

/*
ID:nealgav1
PROG:frac1
LANG:C++
*/
#include<cstdio>
#include<algorithm>
using namespace std;
class node
{public:
int x,y;
float num;
}root[123456];
int Euclid(int a,int b)
{
if(b==0)return a;
Euclid(b,a%b);
}
bool cmp(node a,node b)
{
return a.num<b.num;
}
int main()
{freopen("frac1.in","r",stdin);
freopen("frac1.out","w",stdout);
int m,n,km,nil;
scanf("%d",&m);
km=m;
printf("0/1\n");
nil=0;
for(int j=1;j<m;j++,km--)
for(int i=1;i<km;i++)
{
n=Euclid(i,km);
root[++nil].x=i/n;root[nil].y=km/n;root[nil].num=float(i)/float(km);
}
sort(root+1,root+nil+1,cmp);
km=m;nil=1;
for(int j=1;j<m;j++,km--)
for(int i=1;i<km;i++,nil++)
if(root[nil].x==root[nil-1].x&&root[nil].y==root[nil-1].y)
continue;
else
printf("%d/%d\n",root[nil].x,root[nil].y);
printf("1/1\n");
}

USER: Neal Gavin Gavin [nealgav1]
TASK: frac1
LANG: C++

Compiling...
Compile: OK

Executing...
Test 1: TEST OK [0.011 secs, 4792 KB]
Test 2: TEST OK [0.011 secs, 4792 KB]
Test 3: TEST OK [0.000 secs, 4792 KB]
Test 4: TEST OK [0.000 secs, 4792 KB]
Test 5: TEST OK [0.000 secs, 4792 KB]
Test 6: TEST OK [0.000 secs, 4792 KB]
Test 7: TEST OK [0.000 secs, 4792 KB]
Test 8: TEST OK [0.011 secs, 4792 KB]
Test 9: TEST OK [0.011 secs, 4792 KB]
Test 10: TEST OK [0.011 secs, 4792 KB]
Test 11: TEST OK [0.011 secs, 4792 KB]

All tests OK.
YOUR PROGRAM ('frac1') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.

Here are the test data inputs:

------- test 1 ----
1
------- test 2 ----
2
------- test 3 ----
4
------- test 4 ----
7
------- test 5 ----
10
------- test 6 ----
15------- test 7 ----
24
------- test 8 ----
50
------- test 9 ----
75------- test 10 ----
100
------- test 11 ----
160

Keep up the good work!

Thanks for your submission!

Ordered FractionsRuss Cox

Here's a very fast, straightforward solution from Alex Schwedner:

#include <fstream.h>
#include <stdlib.h>

struct fraction {
int numerator;
int denominator;
};

bool rprime(int a, int b){
int r = a % b;
while(r != 0){
a = b;
b = r;
r = a % b;
}
return(b == 1);
}

int fraccompare (struct fraction *p, struct fraction *q) {
return p->numerator * q->denominator - p->denominator *q->numerator;
}

int main(){
int found = 0;
struct fraction fract[25600];

ifstream filein("frac1.in");
int n;
filein >> n;
filein.close();

for(int bot = 1; bot <= n; ++bot){
for(int top = 0; top <= bot; ++top){
if(rprime(top,bot)){
fract[found].numerator = top;
fract[found++].denominator = bot;
}
}
}

qsort(fract, found, sizeof (struct fraction), fraccompare);

ofstream fileout("frac1.out");
for(int i = 0; i < found; ++i)
fileout << fract[i].numerator << '/' << fract[i].denominator << endl;
fileout.close();

exit (0);
}

Here's a super fast solution from Russ:

We notice that we can start with 0/1 and 1/1 as our ``endpoints'' and recursively generate the middle points by adding numerators and denominators.

0/1                                                              1/1
1/2
1/3                      2/3
1/4              2/5         3/5                 3/4
1/5      2/7     3/8    3/7   4/7   5/8       5/7         4/5

Each fraction is created from the one up to its right and the one up to its left. This idea lends itself easily to a recursion that we cut off when we go too deep.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

int n;
FILE *fout;

/* print the fractions of denominator <= n between n1/d1 and n2/d2 */
void
genfrac(int n1, int d1, int n2, int d2)
{
if(d1+d2 > n)	/* cut off recursion */
return;

genfrac(n1,d1, n1+n2,d1+d2);
fprintf(fout, "%d/%d\n", n1+n2, d1+d2);
genfrac(n1+n2,d1+d2, n2,d2);
}

void
main(void)
{
FILE *fin;

fin = fopen("frac1.in", "r");
fout = fopen("frac1.out", "w");
assert(fin != NULL && fout != NULL);

fscanf(fin, "%d", &n);

fprintf(fout, "0/1\n");
genfrac(0,1, 1,1);
fprintf(fout, "1/1\n");
}