Anti-prime Sequences
| Time Limit: ?3000MS | ? | Memory Limit: ?30000K |
| Total Submissions: ?2175 | ? | Accepted: ?1022 |
Description
Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number. For example, if n = 1 and m = 10, one such anti-prime sequence is 1,3,5,4,2,6,9,7,8,10. This is also the lexicographically first such sequence.?
We can extend the definition by defining a degree danti-prime sequence as one where all consecutive subsequences of length 2,3,...,d sum to a composite number. The sequence above is a degree 2 anti-prime sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11. The lexicographically .rst degree 3 anti-prime sequence for these numbers is 1,3,5,4,6,2,10,8,7,9.?
We can extend the definition by defining a degree danti-prime sequence as one where all consecutive subsequences of length 2,3,...,d sum to a composite number. The sequence above is a degree 2 anti-prime sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11. The lexicographically .rst degree 3 anti-prime sequence for these numbers is 1,3,5,4,6,2,10,8,7,9.?
Input
Input will consist of multiple input sets. Each set will consist of three integers, n, m, and d on a single line. The values of n, m and d will satisfy 1 <= n < m <= 1000, and 2 <= d <= 10. The line 0 0 0 will indicate end of input and should not be processed.
Output
For each input set, output a single line consisting of a comma-separated list of integers forming a degree danti-prime sequence (do not insert any spaces and do not split the output over multiple lines). In the case where more than one anti-prime sequence exists, print the lexicographically first one (i.e., output the one with the lowest first value; in case of a tie, the lowest second value, etc.). In the case where no anti-prime sequence exists, output?
No anti-prime sequence exists.?
No anti-prime sequence exists.?
Sample Input
1 10 2
1 10 3
1 10 5
40 60 7
0 0 0
Sample Output
1,3,5,4,2,6,9,7,8,10
1,3,5,4,6,2,10,8,7,9
No anti-prime sequence exists.
40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54
題意：求n到m的數中任意連續2到d的數的和是合數
DFS
#include<stdio.h>
#include
<
string
.h>
#include
<math.h>
const
int
MAXN=
10001
;
int
n,m,d;
int
vis[MAXN];
int
num[MAXN];
int
pri[MAXN];
int
ans;
int
flag;
void
init ()
{
memset(pri,
0
,
sizeof
(pri));
pri[
0
] = pri[
1
] =
1
;
for
(
int
i =
2
; i <=
100
; i++
)
{
if
( pri[i] )
continue
;
for
(
int
j =
2
; i * j <
10001
; j++
)//這里錯了
pri[i
*j] =
1
;
}
}
bool
judge(
int
t,
int
step)
{
num[step]
=
t;
if
(step>
0
)
{
ans
=
0
;
int
judge=
0
;
for
(
int
j=step; j>=step-d+
1
; j--
)
{
ans
+=
num[j];
if
(judge==
1
&& pri[ans]==
0
)
{
return
false
;
}
judge
=
1
;
if
(j==
0
)
break
;
}
}
return
true
;
}
void
DFS(
int
step)
{
int
ans,i,t;
if
(flag)
return
;
if
(step==m-n+
1
)
//
已經找到了
{
flag
=
1
;
return
;
}
for
(i=n; i<=m; i++
)
{
t
=
0
;
if
(flag)
return
;
if
(!vis[i] &&
judge(i,step))
{
vis[i]
=
1
;
DFS(step
+
1
);
vis[i]
=
0
;
}
}
}
int
main()
{
int
i,j;
init();
while
(scanf(
"
%d%d%d
"
,&n,&m,&
d))
{
flag
=
0
;
if
(n==
0
&& m==
0
&& d==
0
)
break
;
memset(vis,
0
,
sizeof
(vis));
DFS(
0
);
if
(!flag) printf(
"
No anti-prime sequence exists.
"
);
else
{
for
(i=
0
; i<=m-n; i++
)
if
(i==
0
) printf(
"
%d
"
,num[i]);
else
printf(
"
,%d
"
,num[i]);
}
printf(
"
\n
"
);
}
return
0
;
}
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