大数模板
大数加法
void
jial( )
{
memset( sun,
0
,
sizeof
( sun ) );
int
len1
=
strlen( str1 ),len2
=
strlen( str2 );
for
(
int
i
=
0
; i
<
( len1
>
len2
?
len1 : len2 );
++
i )
//
转换成ASCII码
str1[i]
-=
'
0
'
,str2[i]
-=
'
0
'
;
for
(
int
p
=
0
, q
=
len1
-
1
; p
<
q;
++
p ,
--
q )
//
逆序
{
char
c
=
str1[q];
str1[q]
=
str1[p];
str1[p]
=
c;
}
for
(
int
p
=
0
, q
=
len2
-
1
; p
<
q;
++
p,
--
q)
{
char
c
=
str2[q];
str2[q]
=
str2[p];
str2[p]
=
c;
}
for
(
int
i
=
0
,c
=
0
; i
<
( len1
>
len2
?
len1 : len2 )
||
c;
++
i )
//
相加 进位
{
if
( i
<
len1 )
c
+=
str1[i];
if
( i
<
len2 )
c
+=
str2[i];
sun[i]
=
c
%
10
;
//
不同的地方1
c
/=
10
;
//
不同的地方2
}
}
大数乘法
1
#include
"
stdio.h
"
2
#define
MAX 21
3
4
void
InputNumber(
int
[]);
//
输入大数
5
void
InitNumber(
int
[]);
//
初始化用于存放结果的数组
6
void
Multi(
int
[],
int
[],
int
[]);
//
进行大数乘法
7
void
singleMulti(
int
[],
int
,
int
[]);
//
对大数乘以一位数的函数
8
void
Add(
int
[],
int
[]);
//
对结果进行错位相加
9
void
printresult(
int
[]);
//
输出结果
10
11
main()
12
{
13
int
largenumber1[MAX],largenumber2[MAX],mResult[
2
*
MAX
-
1
];
14
//
largenumber1[]和largenumber2[]表示要相乘的大数
15
//
mResult[]表示结果存放的数组，显然其位数不会超过2 * (MAX - 1)
16
printf(
"
Input the first number:\n
"
);
17
InputNumber(largenumber1);
18
printf(
"
Input the second number:\n
"
);
19
InputNumber(largenumber2);
20
InitNumber(mResult);
21
Multi(largenumber1,largenumber2,mResult);
22
printf(
"
The result is:\n
"
);
23
printresult(mResult);
24
getchar();
25
}
26
27
void
InputNumber(
int
largenumber[])
28
{
29
char
numberchar;
30
int
i
=
0
;
31
scanf(
"
%c
"
,
&
numberchar);
32
while
(numberchar
!=
'
\n
'
&&
i
<
MAX)
//
最初输入大数用字符串表示
33
{
34
largenumber[
++
i]
=
numberchar
-
'
0
'
;
35
scanf(
"
%c
"
,
&
numberchar);
36
}
37
largenumber[
0
]
=
i;
38
}
39
40
void
InitNumber(
int
result[])
41
{
42
for
(
int
i
=
0
; i
<
2
*
MAX
-
1
; i
++
)
43
result[i]
=
-
1
;
44
}
45
46
void
Multi(
int
number1[],
int
number2[],
int
result[])
47
{
48
int
i;
49
int
temper[
2
*
MAX
-
1
];
50
for
(i
=
number2[
0
]; i
>=
1
; i
--
)
51
{
52
singleMulti(number1,number2[i],temper);
53
Add(result,temper);
54
}
55
}
56
57
void
singleMulti(
int
number1[],
int
number2,
int
temper[])
58
{
59
int
i,t
=
0
,c
=
2
*
MAX
-
2
;
60
for
(i
=
number1[
0
]; i
>=
1
; i
--
)
61
{
62
int
tempernumber
=
number2
*
number1[i];
63
temper[c
--
]
=
(tempernumber
+
t)
%
10
;
64
t
=
(
int
)((tempernumber
+
t)
/
10
);
65
}
66
if
(t
>
0
)
67
{
68
temper[c]
=
t;
69
temper[
0
]
=
2
*
MAX
-
2
-
c
+
1
;
70
}
71
else
72
temper[
0
]
=
2
*
MAX
-
2
-
c;
73
}
74
75
void
Add(
int
result[],
int
temper[])
76
{
77
static
int
pos
=
2
*
MAX
-
2
;
78
int
c
=
pos,t
=
0
;
79
for
(
int
i
=
2
*
MAX
-
2
; i
>
2
*
MAX
-
2
-
temper[
0
]; i
--
)
80
{
81
if
(result[c]
==
-
1
)
82
result[c]
=
0
;
83
int
tempernumber
=
result[c]
+
temper[i];
84
result[c
--
]
=
(tempernumber
+
t)
%
10
;
85
t
=
(
int
)(tempernumber
/
10
);
86
}
87
if
(t
==
1
)
88
result[c]
=
1
;
89
pos
--
;
90
}
91
92
void
printresult(
int
result[])
93
{
94
for
(
int
i
=
1
; i
<
2
*
MAX
-
1
; i
++
)
95
if
(result[i]
!=
-
1
)
96
printf(
"
%d
"
,result[i]);
97
}