POJ 1416 Shredding Company (dfs)
题意：给定一个目标数n1，一个操作数n2，需要对操作数进行切割，并且各个和要小于目标数，而且最大。
由于操作数n2不大,dfs就可以实现，就是小细节很多，需要注意，对0的判断要加上。
#include
<
iostream
>
#include
<
cstdio
>
#include
<
algorithm
>
#include
<
memory.h
>
#include
<
cmath
>
#include
<
vector
>
using
namespace
std;
const
int
BORDER
=
(
1
<<
26
)
-
1
;
#define
MAXN 1000000
#define
INF 0x7ffffff
#define
_clr(x,y) memset(x,y,sizeof(x))
#define
_add(x) ((++x)&BORDER)
#define
_in(x) scanf("%d",&x)
#define
_out(x) printf("%d\n",x)
#define
_min(m,v) (m)<(v)?(m):(v)
#define
_max(m,v) (m)>(v)?(m):(v)
int
n,target,arr[
10
],sum,mmax,tar_len,t_sum,n_index;
int
pre[
10
],t_pre[
10
],cnt[MAXN],arr_sum[
12
];
int
init()
{
_clr(pre,
0
);
_clr(cnt,
0
);
_clr(arr,
0
);
_clr(t_pre,
0
);
mmax
=
-
1
;
arr[
0
]
=
0
;
target
=
0
;
return
0
;
}
int
dfs(
const
int
&
ilen,
const
int
&
cur_pos,
const
int
&
index)
{
int
i,j,len,tmp;
t_pre[index]
=
ilen;
t_sum
=
0
;
for
(i
=
ilen
-
1
; i
>=
0
;
--
i)
t_sum
=
t_sum
*
10
+
arr[cur_pos
-
i];
if
(
!
index)
arr_sum[index]
=
t_sum;
else
arr_sum[index]
=
arr_sum[index
-
1
]
+
t_sum;
tmp
=
arr_sum[index];
if
( tmp
>
target)
return
0
;
if
(cur_pos
==
n)
{
if
( tmp
==
mmax )
cnt[tmp]
=
1
;
else
if
(tmp
>
mmax)
{
mmax
=
tmp;
n_index
=
index;
memcpy(pre,t_pre,
sizeof
(pre));
}
}
len
=
n
-
cur_pos;
for
(i
=
1
; i
<=
len;
++
i)
dfs(i,cur_pos
+
i,index
+
1
);
return
1
;
}
int
main()
{
char
str1[
10
],str2[
10
];
int
i,j,tmp;
while
(scanf(
"
%s %s
"
,str1,str2))
{
init();
if
(str1[
0
]
==
'
0
'
&&
str2[
0
]
==
'
0
'
)
break
;
tar_len
=
strlen(str1);
for
(i
=
1
; i
<=
tar_len;
++
i)
target
=
target
*
10
+
str1[i
-
1
]
-
'
0
'
;
n
=
strlen(str2);
for
(i
=
1
; i
<=
n;
++
i)
arr[i]
=
str2[i
-
1
]
-
'
0
'
;
dfs(
0
,
0
,
0
);
if
(mmax
==
-
1
)
printf(
"
error\n
"
);
else
if
(cnt[mmax])
printf(
"
rejected\n
"
);
else
{
printf(
"
%d
"
,mmax);
int
cur
=
1
;
for
(i
=
1
;i
<=
n_index;
++
i)
{
tmp
=
pre[i];
printf(
"
"
);
for
(j
=
0
; j
<
tmp;
++
j)
{
printf(
"
%c
"
,arr[cur]
+
'
0
'
);
++
cur;
}
}
printf(
"
\n
"
);
}
}
return
0
;
}