hdu1042(好大的数啊！)
http://acm.hdu.edu.cn/showproblem.php?pid=1402
我的代码，由于时间的问题，以后找机会优化了！（超时）
View Code
#include
"
iostream
"
#define
M 100010
using
namespace
std;
char
ch1[M],ch2[M];
int
a[M],b[M];
int
c[M],d[M];
int
num[M];
int
H[M];
int
t,k,g,v;
int
i,j;
int
La,Lb;
int
sign
=
0
;
void
Add(
int
x ,
int
y)
{
sign
=
0
; t
=
0
;
int
p,q,flag
=
0
;
if
(x
>
y)
{
for
(p
=
0
,q
=
0
;p
<
y,q
<
y; p
++
,q
++
)
{
num[t
++
]
=
(d[p]
+
H[q]
+
flag)
%
10
;
flag
=
(d[p]
+
H[q]
+
flag)
%
10
;
}
for
(p
=
y; p
<
x ; p
++
)
{
num[t
++
]
=
(d[p]
+
flag)
%
10
;
flag
=
(d[p]
+
flag)
/
10
;
}
}
else
if
(x
==
y)
{
for
(p
=
0
,q
=
0
;p
<
x,q
<
y; p
++
,q
++
)
{
num[t
++
]
=
(d[p]
+
H[q]
+
flag)
%
10
;
flag
=
(d[p]
+
H[q]
+
flag)
/
10
;
}
}
else
{
flag
=
0
;
for
(p
=
0
,q
=
0
;p
<
x,q
<
x; p
++
,q
++
)
{
num[t
++
]
=
(H[p]
+
d[q]
+
flag)
%
10
;
flag
=
(H[p]
+
d[q]
+
flag)
/
10
;
}
for
(q
=
x;q
<
y;q
++
)
{
num[t
++
]
=
(H[q]
+
flag)
%
10
;
flag
=
(H[q]
+
flag)
/
10
;
}
}
while
(flag)
{
num[t
++
]
=
flag
%
10
;
flag
/=
10
;
}
for
(
int
yy
=
0
; yy
<
t;yy
++
)
{
d[yy]
=
num[yy];
if
(d[yy]
==
0
) sign
++
;
//
sign用于标记都为结果0的情况
}
g
=
t;
memset(H ,
0
,
sizeof
(H));
memset(num,
0
,
sizeof
(num));
}
void
BigN()
{
int
flag
=
0
;
int
mark
=
0
;
if
(La
>
Lb)
{
for
(i
=
Lb
-
1
;i
>=
0
;i
--
)
{
k
=
0
;
for
(j
=
La
-
1
;j
>=
0
;j
--
)
{
c[k
++
]
=
(b[i]
*
a[j]
+
flag)
%
10
;
flag
=
(b[i]
*
a[j]
+
flag)
/
10
;
}
while
(flag)
{
c[k
++
]
=
flag
%
10
;
flag
/=
10
;
}
//
末尾赋值0
v
=
0
;
for
(
int
xx
=
0
; xx
<
mark; xx
++
)
H[v
++
]
=
0
;
for
(
int
m
=
0
;m
<
k;m
++
)
H[v
++
]
=
c[m];
memset(c,
0
,
sizeof
(c));
//
****************************
Add(g , v);
mark
++
;
}
}
else
{
for
(i
=
La
-
1
;i
>=
0
;i
--
)
{
k
=
0
;
for
(j
=
Lb
-
1
;j
>=
0
;j
--
)
{
c[k
++
]
=
(a[i]
*
b[j]
+
flag)
%
10
;
flag
=
(a[i]
*
b[j]
+
flag)
/
10
;
}
while
(flag)
{
c[k
++
]
=
flag
%
10
;
flag
/=
10
;
}
//
末尾赋值0
v
=
0
;
for
(
int
xx
=
0
; xx
<
mark; xx
++
)
H[v
++
]
=
0
;
for
(
int
m
=
0
;m
<
k;m
++
)
H[v
++
]
=
c[m];
//
****************************
Add(g , v);
mark
++
;
}
}
}
int
main()
{
while
(cin
>>
ch1
>>
ch2)
{
t
=
0
;k
=
0
;g
=
0
;
La
=
strlen(ch1);
Lb
=
strlen(ch2);
for
(i
=
0
;i
<
La;i
++
) a[i]
=
ch1[i]
-
'
0
'
;
for
(i
=
0
;i
<
Lb;i
++
) b[i]
=
ch2[i]
-
'
0
'
;
BigN();
if
(sign
!=
t)
{
for
(i
=
t
-
1
;i
>=
0
;i
--
)
cout
<<
d[i];
cout
<<
endl;
}
else
cout
<<
"
0
"
<<
endl;
}
return
0
;
}
一位dn的，神人啊！
http://www.cppblog.com/misschuer/archive/2010/01/16/80356.html
View Code
#include
<
iostream
>
#include
<
cmath
>
using
namespace
std;
typedef
struct
vir{
double
re,im;
vir(){}
vir(
double
a,
double
b){re
=
a;im
=
b;}
vir
operator
+
(
const
vir
&
b){
return
vir(re
+
b.re,im
+
b.im);}
vir
operator
-
(
const
vir
&
b){
return
vir(re
-
b.re,im
-
b.im);}
vir
operator
*
(
const
vir
&
b){
return
vir(re
*
b.re
-
im
*
b.im,re
*
b.im
+
b.re
*
im);}
}vir;
vir x1[
200005
],x2[
200005
];
const
double
Pi
=
acos(
-
1.0
);
void
change(vir
*
x,
int
len,
int
loglen)
{
int
i,j,k,t;
for
(i
=
0
;i
<
len;i
++
)
{
t
=
i;
for
(j
=
k
=
0
;j
<
loglen;j
++
,t
>>=
1
)
k
=
(k
<<
1
)
|
(t
&
1
);
if
(k
<
i)
{
vir wt
=
x[k];
x[k]
=
x[i];
x[i]
=
wt;
}
}
}
void
fft(vir
*
x,
int
len,
int
loglen)
{
int
i,j,t,s,e;
change(x,len,loglen);
t
=
1
;
for
(i
=
0
;i
<
loglen;i
++
,t
<<=
1
)
{
s
=
0
;
e
=
s
+
t;
while
(s
<
len)
{
vir a,b,wo(cos(Pi
/
t),sin(Pi
/
t)),wn(
1
,
0
);
for
(j
=
s;j
<
s
+
t;j
++
)
{
a
=
x[j];
b
=
x[j
+
t]
*
wn;
x[j]
=
a
+
b;
x[j
+
t]
=
a
-
b;
wn
=
wn
*
wo;
}
s
=
e
+
t;
e
=
s
+
t;
}
}
}
void
dit_fft(vir
*
x,
int
len,
int
loglen)
{
int
i,j,s,e,t
=
1
<<
loglen;
for
(i
=
0
;i
<
loglen;i
++
)
{
t
>>=
1
;
s
=
0
;
e
=
s
+
t;
while
(s
<
len)
{
vir a,b,wn(
1
,
0
),wo(cos(Pi
/
t),
-
sin(Pi
/
t));
for
(j
=
s;j
<
s
+
t;j
++
)
{
a
=
x[j]
+
x[j
+
t];
b
=
(x[j]
-
x[j
+
t])
*
wn;
x[j]
=
a;
x[j
+
t]
=
b;
wn
=
wn
*
wo;
}
s
=
e
+
t;
e
=
s
+
t;
}
}
change(x,len,loglen);
for
(i
=
0
;i
<
len;i
++
)
x[i].re
/=
len;
}
int
main()
{
char
a[
100005
],b[
100005
];
int
i,len1,len2,t,over,len,loglen;
while
(scanf(
"
%s%s
"
,a,b)
!=
EOF)
{
len1
=
strlen(a)
<<
1
;
len2
=
strlen(b)
<<
1
;
len
=
1
;
loglen
=
0
;
while
(len
<
len1)
{
len
<<=
1
;
loglen
++
;
}
while
(len
<
len2)
{
len
<<=
1
;
loglen
++
;
}
for
(i
=
0
;a[i]
!=
'
\0
'
;i
++
)
{
x1[i].re
=
a[i]
-
'
0
'
;
x1[i].im
=
0
;
}
for
(;i
<
len;i
++
)
x1[i].re
=
x1[i].im
=
0
;
for
(i
=
0
;b[i]
!=
'
\0
'
;i
++
)
{
x2[i].re
=
b[i]
-
'
0
'
;
x2[i].im
=
0
;
}
for
(;i
<
len;i
++
)
x2[i].re
=
x2[i].im
=
0
;
fft(x1,len,loglen);
fft(x2,len,loglen);
for
(i
=
0
;i
<
len;i
++
)
x1[i]
=
x1[i]
*
x2[i];
dit_fft(x1,len,loglen);
for
(i
=
(len1
+
len2)
/
2
-
2
,over
=
loglen
=
0
;i
>=
0
;i
--
)
{
t
=
x1[i].re
+
over
+
0.5
;
a[loglen
++
]
=
t
%
10
;
over
=
t
/
10
;
}
while
(over)
{
a[loglen
++
]
=
over
%
10
;
over
/=
10
;
}
for
(loglen
--
;loglen
>=
0
&&!
a[loglen];loglen
--
);
if
(loglen
<
0
)
putchar(
'
0
'
);
else
for
(;loglen
>=
0
;loglen
--
)
putchar(a[loglen]
+
'
0
'
);
putchar(
'
\n
'
);
}
return
0
;
}