POJ 1837 Balance
解题思路:DP
g[i]表示第i个砝码的重量，c[j]表示第j个点的坐标，opt[i][k]表示加了i个砝码，力矩之和为k的方法数
DP公式为(考虑到力矩为负，因此需要做偏移):
opt[i
+
1
][k
+
g[i]
*
c[j]
+=
opt[i][k]
第i次迭代只于i-1次有关，因此可以缩减为二维数组
NULL
#include
<
iostream
>
using
namespace
std;
#define
MID 4875
#define
MAXN 9751
int
main()
{
int
i,j,k,m,n,t,
*
x,
*
y,
*
z,c[
20
],g[
20
],a[MAXN],b[MAXN];
scanf(
"
%d %d
"
,
&
m,
&
n);
for
(i
=
0
;i
<
m;i
++
)scanf(
"
%d
"
,
&
c[i]);
for
(i
=
0
;i
<
n;i
++
)scanf(
"
%d
"
,
&
g[i]);
memset(a,
0
,
sizeof
(a));
a[MID]
=
1
,x
=
a,y
=
b;
for
(i
=
0
;i
<
n;i
++
)
{
for
(j
=
0
;j
<
MAXN;j
++
)y[j]
=
0
;
for
(j
=
0
;j
<
m;j
++
)
for
(t
=
g[i]
*
c[j],k
=
0
;k
<
MAXN;k
++
)
if
(x[k])y[k
+
t]
+=
x[k];
z
=
x,x
=
y,y
=
z;
}
printf(
"
%d\n
"
,x[MID]);
return
0
;
}