Exponentiation--POJ 1001
1、题目类型：数论、大数乘法。
2、解题思路：二分计算提高运算速度：n表示为2的0、1、2、3...次幂的之和。
3、注意事项：测试数据中00.00的情况；注意 . 的位置。
4、实现方法:(C++)
#include
<
iostream
>
#include
<
string
>
#include
<
algorithm
>
using
namespace
std;
string
ans;
string
Solve(
string
str1,
string
str2)
{
int
len1
=
str1.length();
int
len2
=
str2.length();
int
i,j,a,b;
string
str
=
""
;
int
tmp[
1000
];
reverse(str1.begin(),str1.end());
reverse(str2.begin(),str2.end());
memset(tmp,
0
,
sizeof
(tmp));
for
(i
=
0
;i
<
len1;i
++
)
{
for
(j
=
0
;j
<
len2;j
++
)
{
a
=
str1[i]
-
'
0
'
;
b
=
str2[j]
-
'
0
'
;
tmp[i
+
j]
+=
a
*
b;
}
}
for
(i
=
0
;i
<
len1
+
len2
-
1
;i
++
)
{
a
=
tmp[i]
%
10
;
b
=
tmp[i]
/
10
;
tmp[i]
=
a;
tmp[i
+
1
]
+=
b;
}
if
(b
>
0
)
i
++
;
for
(j
=
0
;j
<
i;j
++
)
str
+=
tmp[j]
+
'
0
'
;
reverse(str.begin(),str.end());
return
str;
}
void
Cal(
string
str,
int
n)
{
ans
=
"
1
"
;
while
(n)
{
if
(n
&
1
)
ans
=
Solve(ans,str);
str
=
Solve(str,str);
n
=
n
>>
1
;
}
}
int
main()
{
string
str;
int
i,n,pos,step
=
0
;
while
(cin
>>
str
>>
n)
{
pos
=-
1
;
for
(i
=
0
;i
<
str.length();i
++
)
{
if
(str[i]
==
'
.
'
)
{
pos
=
i;
break
;
}
}
if
(pos
!=-
1
)
{
step
=
str.length()
-
pos
-
1
;
str.erase(str.begin()
+
pos);
}
for
(i
=
0
;i
<
str.length();i
++
)
{
if
(str[i]
!=
'
0
'
)
break
;
}
if
(i
==
str.length())
{
cout
<<
0
<<
endl;
continue
;
}
Cal(str,n);
string
s
=
""
;
int
flag
=
0
;
int
flag1
=
0
;
reverse(ans.begin(),ans.end());
for
(i
=
0
;i
<
ans.length();i
++
)
{
if
(
!
flag1)
{
if
((step
&&
i
==
step
*
n))
{
if
(flag)
s
+=
'
.
'
;
flag1
=
1
;
}
}
if
(
!
flag
&&!
flag1)
{
if
(ans[i]
!=
'
0
'
)
{
s
+=
ans[i];
flag
=
1
;
}
}
else
{
s
+=
ans[i];
}
}
if
(
!
flag1
&&
i
<=
step
*
n)
s
+=
'
.
'
;
reverse(s.begin(),s.end());
while
(s[
0
]
==
'
0
'
)
{
s.erase(s.begin());
}
cout
<<
s
<<
endl;
}
return
0
;
}
5、实现方法：（java）
import
java.util.
*
;
import
java.math.
*
;
public
class
Main
{
public
static
void
main(String[] args)
{
Solve slv
=
new
Solve();
slv.Do();
}
}
class
Solve
{
Scanner cin
=
new
Scanner(System.in);
String a,s;
int
n,start,len,tmp;
void
Do()
{
while
(cin.hasNext())
{
boolean
flag
=
true
;
a
=
cin.next();
n
=
cin.nextInt();
Double dd
=
new
Double(a);
if
(dd.intValue()
==
dd.doubleValue())
flag
=
false
;
BigDecimal bigd
=
new
BigDecimal(a);
s
=
bigd.pow(n).toPlainString();
if
(flag
==
true
)
{
start
=
0
;
if
(s.charAt(
0
)
==
'
0
'
&&
s.charAt(
1
)
==
'
.
'
)
start
=
1
;
len
=
s.length()
-
1
;
while
(len
>=
0
&&
s.charAt(len)
==
'
0
'
)
len
--
;
s
=
s.substring(start,len
+
1
);
}
else
{
len
=
s.length()
-
1
;
while
(len
>=
0
&&
s.charAt(len)
!=
'
.
'
)
len
--
;
if
(len
>=
0
)
s
=
s.substring(
0
,len);
}
if
(s.length()
==
0
)
s
=
"
0
"
;
System.out.println(s);
}
}
}