ACM@BIT Another A+B Problem
Description
Calculate a+b
Input
Two integer a,b;(-10^1000<=a,b<=10^1000)
Output
Output a+b
Sample Input
1 2
Sample Output
3
1
#include
<
iostream
>
2
#include
<
string
>
3
#include
<
cstring
>
4
using
namespace
std;
5
6
const
int
MAXNUM
=
1002
;
7
class
BigNum {
8
public
:
9
BigNum();
10
BigNum(
const
string
&
);
11
BigNum
operator
+
(BigNum
&
);
12
void
print()
const
;
13
void
setLen(
int
);
14
private
:
15
char
num[MAXNUM];
16
int
len;
17
int
flag;
//
标记正负 ,0为正
18
};
19
20
BigNum::BigNum() {
21
22
}
23
24
BigNum::BigNum(
const
string
&
param) {
25
strcpy(num, param.c_str());
26
len
=
param.length();
27
for
(
int
i
=
0
; i
<
len
/
2
; i
++
) {
28
char
a
=
num[len
-
i
-
1
];
29
num[len
-
i
-
1
]
=
num[i];
30
num[i]
=
a;
31
}
32
if
(num[len
-
1
]
==
'
-
'
) {
33
flag
=
1
;
34
}
else
{
35
flag
=
0
;
36
}
37
}
38
39
BigNum BigNum::
operator
+
(BigNum
&
param) {
40
BigNum sum;
41
42
int
maxLen
=
(len
-
flag)
>
(param.len
-
param.flag)
?
(len
-
flag)
43
: (param.len
-
param.flag);
44
setLen(maxLen);
45
param.setLen(maxLen);
//
调整长度,使数字部分对齐
46
47
if
(flag
==
param.flag) {
48
int
singal
=
0
;
//
进位
49
int
i
=
0
;
50
for
(; i
<
maxLen; i
++
) {
51
sum.num[i]
=
(num[i]
-
96
+
param.num[i]
+
singal)
%
10
+
48
;
52
if
(num[i]
+
param.num[i]
-
96
+
singal
>
9
) {
53
singal
=
1
;
54
}
else
{
55
singal
=
0
;
56
}
//
设置进位
57
}
58
if
(singal
==
1
) {
59
sum.num[i]
=
'
1
'
;
60
i
++
;
61
}
//
最高位进位
62
if
(flag
==
1
) {
63
sum.num[i]
=
'
-
'
;
64
i
++
;
65
}
//
结果的符号
66
sum.len
=
i;
67
sum.flag
=
flag;
68
}
else
{
69
int
flag1
=
0
;
//
标记绝对值大小
70
for
(
int
i
=
maxLen; i
>
0
; i
--
) {
71
if
(num[i
-
1
]
>
param.num[i
-
1
]) {
72
flag1
=
1
;
73
break
;
74
}
75
if
(num[i
-
1
]
<
param.num[i
-
1
]) {
76
flag1
=
-
1
;
77
break
;
78
}
79
}
80
if
(flag1
==
0
) {
81
sum
=
BigNum(
"
0
"
);
82
}
83
if
(flag1
==
1
) {
84
sum.flag
=
flag;
85
int
osingal
=
0
;
//
标记借位
86
int
Len
=
maxLen;
87
for
(
int
i
=
0
; i
<
maxLen; i
++
) {
88
sum.num[i]
=
(num[i]
+
10
-
param.num[i]
-
osingal)
%
10
+
48
;
89
if
(num[i]
-
param.num[i]
-
osingal
<
0
) {
90
osingal
=
1
;
91
}
else
{
92
osingal
=
0
;
93
}
94
}
95
for
(; Len
>
0
; Len
--
) {
96
if
(sum.num[Len
-
1
]
!=
'
0
'
)
97
break
;
98
}
99
if
(sum.flag
==
1
) {
100
sum.num[Len]
=
'
-
'
;
101
Len
++
;
102
}
103
sum.len
=
Len;
104
}
105
if
(flag1
==
-
1
) {
106
sum.flag
=
param.flag;
107
int
osingal
=
0
;
//
标记借位
108
int
Len
=
maxLen;
109
for
(
int
i
=
0
; i
<
maxLen; i
++
) {
110
sum.num[i]
=
(param.num[i]
+
10
-
num[i]
-
osingal)
%
10
+
48
;
111
if
(param.num[i]
-
num[i]
-
osingal
<
0
) {
112
osingal
=
1
;
113
}
else
{
114
osingal
=
0
;
115
}
116
}
117
for
(; Len
>
0
; Len
--
) {
118
if
(sum.num[Len
-
1
]
!=
'
0
'
)
119
break
;
120
}
121
if
(sum.flag
==
1
) {
122
sum.num[Len]
=
'
-
'
;
123
Len
++
;
124
}
125
sum.len
=
Len;
126
}
127
}
128
return
sum;
129
}
130
131
void
BigNum::print()
const
{
132
for
(
int
i
=
len; i
>
0
; i
--
) {
133
cout
<<
num[i
-
1
];
134
}
135
cout
<<
endl;
136
}
137
138
void
BigNum::setLen(
int
param) {
//
调整，使两数除符号位后长度相同
139
if
(flag
==
1
) {
140
num[param]
=
'
-
'
;
141
}
142
for
(
int
i
=
len
-
flag; i
<
param; i
++
) {
143
num[i]
=
'
0
'
;
144
}
145
}
146
147
int
main(
void
) {
148
string
a, b;
149
cin
>>
a
>>
b;
150
if
(a.length()
>
1002
||
b.length()
>
1002
)
151
return
0
;
152
BigNum add1(a), add2(b), temp;
153
temp
=
add1
+
add2;
154
temp.print();
155
return
1
;
156
}
大数加减法