USACO 2.1 Ordered Fractions
很简单的一道题
直接找出所有的分数，然后按值排序，去除重复的输出。
my code
/*
ID: superbi1
LANG: C
TASK: frac1
*/
#include
<
stdio.h
>
#include
<
string
.h
>
#include
<
stdlib.h
>
#include
<
math.h
>
#define
NL 160*161
#define
EP 1e-10
struct
Fct{
int
a, b;
double
val;
}f[NL];
int
nf;
int
gcd(
int
a,
int
b)
{
if
(b
==
0
)
return
a;
return
gcd(b, a
%
b);
}
int
cmp(
const
void
*
a,
const
void
*
b)
{
struct
Fct
*
x
=
(
struct
Fct
*
)a;
struct
Fct
*
y
=
(
struct
Fct
*
)b;
if
(fabs(x
->
val
-
y
->
val)
<
EP)
return
0
;
else
if
(x
->
val
<
y
->
val)
return
-
1
;
else
return
1
;
}
int
cmp1(
struct
Fct
*
a,
struct
Fct
*
b)
{
return
a
->
a
-
a
->
b;
}
int
main()
{
FILE
*
in
=
fopen(
"
frac1.in
"
,
"
r
"
);
FILE
*
out
=
fopen(
"
frac1.out
"
,
"
w
"
);
int
n;
int
I, K;
int
a, b;
fscanf(
in
,
"
%d
"
,
&
n);
nf
=
0
;
f[nf].a
=
0
;
f[nf].b
=
1
;
f[nf
++
].val
=
0
;
f[nf].a
=
1
;
f[nf].b
=
1
;
f[nf
++
].val
=
1
;
for
(I
=
1
; I
<=
n; I
++
) {
for
(K
=
1
; K
<
I; K
++
) {
int
g
=
gcd(I, K);
f[nf].a
=
K
/
g;
f[nf].b
=
I
/
g;
f[nf
++
].val
=
K
*
1.0
/
I;
}
}
qsort(f, nf,
sizeof
(f[
0
]), cmp);
a
=
f[
0
].a;
b
=
f[
0
].b;
for
(I
=
1
; I
<
nf; I
++
) {
if
(a
==
f[I].a
&&
b
==
f[I].b)
continue
;
else
{
fprintf(
out
,
"
%d/%d\n
"
, a, b);
a
=
f[I].a;
b
=
f[I].b;
}
}
fprintf(
out
,
"
%d/%d\n
"
, a, b);
return
0
;
}
改进：
1.发现这些分数的分子和分母互素
2.排序的时候的小技巧：p1/q1 = p2/q2 --> p1*q2 = p2*q1
stand1
#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
);
}
标程二：
改进：
1.找出这些分数之间的规律，然后直接生成
stand2
#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
"
);
}