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
