贴两个水题的解题报告
和几何相关的东西吧……我都不好意思说是几何……
其实就是水题
HOJ1027
我的做法很囧，旋转小角度然后比判断，精度要求比较高，不知到有什么好的方法没有
HOJ 1027
1
#include
<
iostream
>
2
#include
<
cstring
>
3
#include
<
cmath
>
4
#include
<
cstdio
>
5
using
namespace
std;
6
7
const
double
PI
=
acos(
-
1.0
);
8
const
double
D
=
0.002
;
9
/*
10
* === FUNCTION ======================================================================
11
* Name: judge
12
* Description:
13
* =====================================================================================
14
*/
15
bool
16
judge(
double
a,
double
b,
double
x,
double
y)
17
{
18
if
(a
>
x
&&
b
>
y)
19
return
true
;
20
else
if
(a
<=
x)
21
return
false
;
22
for
(
double
i
=
0
;i
<=
PI
/
2
;i
+=
D)
23
{
24
double
tmpy
=
(x
/
2
*
tan(i)
+
y
/
2
)
*
cos(i)
*
2
;
25
double
tmpx
=
(y
/
2
*
tan(i)
+
x
/
2
)
*
cos(i)
*
2
;
26
if
(tmpy
<
b
&&
tmpx
<
a)
27
return
true
;
28
}
29
return
false
;
30
}
/*
----- end of function judge -----
*/
31
32
int
main()
33
{
34
double
a, b, x, y;
35
int
cases;
36
scanf (
"
%d
"
,
&
cases);
37
while
(cases
--
)
38
{
39
scanf(
"
%lf%lf%lf%lf
"
,
&
a,
&
b,
&
x,
&
y);
40
if
(a
>
b) swap(a, b);
41
if
(x
>
y) swap(x, y);
42
puts(judge(a, b, x, y)
?
"
Escape is possible.
"
:
"
Box cannot be dropped.
"
);
43
}
44
return
0
;
45
}
HOJ1032 计算重量分布均匀的多边形的重心
先用第0个点做N - 1个三角形求重心，然后就转移到只有点有重量的离散函数积分，工数上的内容了……
HOJ1032
1
#include
<
iostream
>
2
#include
<
cstring
>
3
#include
<
cmath
>
4
#include
<
cstdio
>
5
using
namespace
std;
6
const
int
N
=
1000000
;
7
const
double
EPS
=
0.001
;
8
double
x[N], y[N];
9
double
multiply(
double
x1,
double
y1,
double
x2,
double
y2)
10
{
11
return
x1
*
y2
-
x2
*
y1;
12
}
13
int
main()
14
{
15
int
cases, n;
16
scanf (
"
%d
"
,
&
cases);
17
while
(cases
--
)
18
{
19
scanf(
"
%d
"
,
&
n);
20
double
s
=
0
, sumsx
=
0
, sumsy
=
0
;
21
double
ds, dx, dy;
22
for
(
int
i
=
0
;i
<
n;i
++
) scanf(
"
%lf %lf
"
,
&
x[i],
&
y[i]);
23
for
(
int
i
=
2
;i
<
n;i
++
)
24
{
25
double
ds
=
multiply(x[i]
-
x[
0
], y[i]
-
y[
0
], x[i
-
1
]
-
x[
0
], y[i
-
1
]
-
y[
0
]);
26
dx
=
(x[
0
]
+
x[i
-
1
]
+
x[i])
/
3
;
27
dy
=
(y[
0
]
+
y[i
-
1
]
+
y[i])
/
3
;
28
s
+=
ds;
29
sumsx
+=
dx
*
ds;
30
sumsy
+=
dy
*
ds;
31
}
32
double
ansx
=
sumsx
/
s, ansy
=
sumsy
/
s;
33
if
(fabs(ansx)
<
EPS)
34
ansx
=
0
;
35
if
(fabs(ansy)
<
EPS)
36
ansy
=
0
;
37
printf(
"
%0.2lf %0.2lf\n
"
, ansx, ansy);
38
}
39
return
0
;
40
}