2970.
Hackle Number
Time Limit: 1.0 Seconds
Memory Limit: 65536K
Total Runs: 183
Accepted Runs: 81
When I was in college one of my chemistry professor invented a machine, which can make a benzin chain. A benzin chain is made by carbon atom and look like following:
Benzin chainBenzin chainBenzin chainwith one cycle with two cycle with three cycle
One of the disadvantages of that machine is it is very costly and if we do not supply it proper number of carbon then it cannot complete its operation and destroy the entire atom. Its need 1000$ for make a benzin chain of any length or unsuccessful operation. Acutely it was very inefficient machine. But he has no way rather than using it. So he asked me to make a program for him that can complete a proper number of carbon and its volume in standard temperature and pressure. I solve it that time. But now I am in reverse situation. He again ask me to extend that program so that if he give me a number of atom I have to find is it possible to build a carbon chain with that given number of atom. Would you please help me?
Input
The input contains a several test cases. Each test case will contain a single integer not more than 100 digit in one line. The EOF indicates the end of file.
Output
There will be one line output for one input. Each of the output line contain one of the two words "Possible." Or "Not possible.". If it is possible to make a carbon chain with the given number M then print "Possible." Otherwise print "Not possible.".
Sample Input
6
10
14
15
100
Sample Output
"Possible."
"Possible."
"Possible."
"Not possible."
"Not possible."
Problem Setter: M H Rasel.
Source: CUET Easy Contest
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//
规律:f(n)=4*n+2,如果满足就是可能的,否则不可能
#include
<
iostream
>
using
namespace
std; typedef
struct
node {
char
num[
105
];
int
len; }Num;
void
Change(Num
&
ob,
int
n) {
int
i;
for
(i
=
0
;i
<
ob.len;i
++
) ob.num[i]
-=
n; }
void
Rev(Num
&
ob) {
int
s,e;
char
temp; s
=
0
; e
=
ob.len
-
1
;
while
(s
<
e) { temp
=
ob.num[s]; ob.num[s]
=
ob.num[e]; ob.num[e]
=
temp; s
++
;
--
e; }
return
; } Num Sub(Num
&
a,Num
&
b) { Num c; memset(
&
c,
0
,
sizeof
(c));
int
tw
=
0
,i,l
=
a.len; c.len
=
a.len;
for
(i
=
0
;i
<=
l;i
++
) { c.num[i]
=
a.num[i]
-
b.num[i]
-
tw;
if
(c.num[i]
<
0
) { tw
=
1
; c.num[i]
+=
10
; }
else
tw
=
0
; }
while
(c.len
>
1
&&!
c.num[c.len
-
1
])
--
c.len;
return
c; }
void
Mult_ten(Num
&
ob) {
int
i;
if
(ob.len
==
1
&&
ob.num[
0
]
==
0
)
return
;
for
(i
=
ob.len;i
>
0
;i
--
) { ob.num[i]
=
ob.num[i
-
1
]; } ob.num[
0
]
=
0
; ob.len
++
; }
int
Cmp(Num
&
a,Num
&
b) { Num c; memset(
&
c,
0
,
sizeof
(c));
if
(a.len
>
b.len)
return
1
;
else
if
(a.len
<
b.len)
return
-
1
;
else
{
int
i;
for
(i
=
a.len
-
1
;i
>=
0
;
--
i) {
if
(a.num[i]
>
b.num[i])
return
1
;
else
if
(a.num[i]
<
b.num[i])
return
-
1
; } }
return
0
; } Num Mod(Num
&
a,Num
&
b) { Num temp; memset(
&
temp,
0
,
sizeof
(temp));
for
(
int
i
=
a.len
-
1
;i
>=
0
;
--
i) { Mult_ten(temp); temp.num[
0
]
=
a.num[i];
while
(Cmp(temp,b)
>=
0
) { temp
=
Sub(temp,b); } }
return
temp; }
int
main() { Num a,b,ans,c; memset(
&
a,
0
,
sizeof
(a)); memset(
&
b,
0
,
sizeof
(b));
while
(scanf(
"
%s
"
,a.num)
!=
EOF) { a.len
=
strlen(a.num); strcpy(b.num,
"
2
"
); b.len
=
1
; Change(a,
'
0
'
); Change(b,
'
0
'
); Rev(a); Rev(b); c
=
Sub(a,b); Change(c,
-
'
0
'
); Rev(c); strcpy(b.num,
"
4
"
); b.len
=
1
; Change(c,
'
0
'
); Change(b,
'
0
'
); Rev(c); Rev(b); ans
=
Mod(c,b); Change(ans,
-
'
0
'
); Rev(ans);
if
(ans.len
==
1
&&
ans.num[
0
]
==
'
0
'
) printf(
"
\
"
Possible.\
"
\n
"
);
else
printf(
"
\
"
Not possible.\
"
\n
"
); memset(
&
a,
0
,
sizeof
(a)); memset(
&
b,
0
,
sizeof
(b)); }
return
0
; }
转载于:https://www.cnblogs.com/forever4444/archive/2009/05/19/1460369.html
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