POJ 1426 Find The Multiple
解题思路:
1）每5位为一个基数，枚举5位所有只包含0,1的数ans[]，map[i][j]保存(ans[j]*100000^i)%n的值
2）BFS计算所有组合出现的模，直到出现0
代码
#include
<
iostream
>
using
namespace
std;
#define
MAXN 201
#define
radix 100000
const
int
eu[]
=
{
1
,
10
,
11
,
100
,
101
,
110
,
111
,
1000
,
1001
,
1010
,
1011
,
1100
,
1101
,
1110
,
1111
,
10000
,
10001
,
10010
,
10011
,
10100
,
10101
,
10110
,
10111
,
11000
,
11001
,
11010
,
11011
,
11100
,
11101
,
11110
,
11111
};
int
main()
{
bool
visit[MAXN];
int
id[MAXN][
2
],mod[
20
][
32
], q[MAXN], next[MAXN], ans[
20
];
int
n, p, i, j, k, t, iter, e;
while
(scanf(
"
%d
"
,
&
n)
&&
n)
{
for
(i
=
0
;i
<
MAXN;i
++
)visit[i]
=
0
,next[i]
=-
1
;
memset(ans,
0
,
sizeof
(ans));
for
(i
=
0
;i
<
20
; i
++
)
{
for
(j
=
0
;j
<
31
;j
++
)
{
if
(i
>
0
)mod[i][j]
=
(mod[i
-
1
][j]
*
radix)
%
n;
else
mod[i][j]
=
eu[j]
%
n;
if
(i
==
0
&&!
visit[mod[i][j]])visit[mod[i][j]]
=
true
,id[mod[i][j]][
0
]
=
i,id[mod[i][j]][
1
]
=
j;
}
}
for
(p
=
i
=
0
;i
<
n;i
++
)
if
(visit[i])q[p
++
]
=
i;
for
(i
=
1
;i
<
20
&&!
visit[
0
];i
++
)
{
for
(e
=
p,j
=
0
;j
<
31
&&!
visit[
0
];j
++
)
{
t
=
mod[i][j];
if
(t
==
0
)
int
b
=
0
;
if
(
!
visit[t])visit[t]
=
true
,id[t][
0
]
=
i,id[t][
1
]
=
j,q[p
++
]
=
t;
for
(k
=
0
;k
<
e
&&!
visit[
0
];k
++
)
{
t
=
(mod[i][j]
+
q[k])
%
n;
if
(t
==
0
)
int
b
=
0
;
if
(
!
visit[t])visit[t]
=
true
,id[t][
0
]
=
i,id[t][
1
]
=
j,next[t]
=
q[k],q[p
++
]
=
t;
}
}
}
for
(iter
=
id[
0
][
0
],i
=
0
;i
!=-
1
;i
=
next[i])
{
j
=
id[i][
0
],k
=
id[i][
1
];
p
=
iter
-
j;
ans[p]
=
eu[k];
}
printf(
"
%d
"
,ans[
0
]);
for
(i
=
1
;i
<=
iter;i
++
)
printf(
"
%05d
"
, ans[i]);
printf(
"
\n
"
);
}
return
0
;
}