POJ 1276 Cash Machine
解题思路:dp(多重背包问题)
将多重背包转化为01背包，假设每个物品的数量上限为n，每件的体积为0，价值为w
那么可以把该物品分为1,2,4,...,2^(k-1)和n-(2^k+1),那么新的物品就是(体积，价值):(0,w),(0,2w),...,(0,2^(k-1)w)和(0,(n-2^k+1)w)
(这样可以构造出任何<=n的情况，假设n为7，则分成系数为1,2和4,那么1-7的所有情况均可由1,2,4这3个基数组合得到)
NULL
#include
<
iostream
>
using
namespace
std;
#define
MAXN 100001
int
main()
{
int
c,m,i,j,k,t,p;
int
d[
11
],n[
11
],dp[MAXN];
while
(scanf(
"
%d %d
"
,
&
c,
&
m)
!=
EOF)
{
for
(i
=
0
;i
<
m;i
++
)scanf(
"
%d %d
"
,
&
n[i],
&
d[i]);
memset(dp,
0
,
sizeof
(dp));
for
(i
=
0
;i
<
m;i
++
)
if
(n[i]
*
d[i]
==
c){dp[c]
=
c;
break
;}
else
if
(n[i]
*
d[i]
>
c)
{
for
(j
=
d[i];j
<=
c;j
++
)
if
(dp[j
-
d[i]]
+
d[i]
>
dp[j])dp[j]
=
dp[j
-
d[i]]
+
d[i];
}
else
{
for
(t
=
n[i],j
=
1
;j
<=
(n[i]
>>
1
);t
-=
j,j
*=
2
)
for
(k
=
c,p
=
j
*
d[i];k
>=
p;k
--
)
if
(dp[k
-
p]
+
p
>
dp[k])dp[k]
=
dp[k
-
p]
+
p;
for
(k
=
c,p
=
t
*
d[i];k
>=
p;k
--
)
if
(dp[k
-
p]
+
p
>
dp[k])dp[k]
=
dp[k
-
p]
+
p;
}
printf(
"
%d\n
"
,dp[c]);
}
return
0
;
}