挑战面试编程：大整数的加、减、乘、除
一切都是有限的，哪怕是看起来无限的时间或空间都非常可能是有限的。在计算机中内置类型的加、减、乘、除都是有限的。我们来实现一个“无限”的大整数加、减、乘、除。
下面使用C++代码实现
#include <iostream>
#include <string>
using namespace std;
//大整数的加减乘除
string add_int(string, string);
string sub_int(string, string);
string mul_int(string, string);
string div_int(string, string);
string mod_int(string, string);
string divide_int(string, string, int);
inline int compare(string s1, string s2)
{
if (s1.size() < s2.size())
return -1;
else if (s1.size() > s2.size())
return 1;
else
return s1.compare(s2);
}
/*
大整数加法
本质上仅仅处理：两个正数相加，如 "123" + "234"
其他情况需转化
1. 正加负 => "123" + "-234" = "123" - "234" 转化为减法
2. 负加正 => "-234" + "123" = "123" - "234"
3. 负加负 => "-123" + "-234" = -("123" + "234")
*/
string add_int(string s1, string s2)
{
if (s1 == "0")
return s2;
if (s2 == "0")
return s1;
if (s1[0] == '-')
{
if (s2[0] == '-')
{
return "-" + add_int(s1.erase(0, 1), s2.erase(0, 1)); //情况三
}
else
{
return sub_int(s2, s1.erase(0, 1)); //情况二
}
}
if (s2[0] == '-')
{
return sub_int(s1, s2.erase(0, 1)); //情况一
}
//处理本质情况
string::size_type i, size1, size2;
size1 = s1.size();
size2 = s2.size();
if (size1 < size2)
{
for (i = 0; i < size2 - size1; i++) //在s1左边补零
s1 = "0" + s1;
}
else
{
for (i = 0; i < size1 - size2; i++) //在s2左边补零
s2 = "0" + s2;
}
int n1, n2;
n2 = 0;
size1 = s1.size();
size2 = s2.size();
string res;
for (i = size1 - 1; i != 0; i--) //从最低位加起
{
n1 = (s1[i] - '0' + s2[i] - '0' + n2) % 10; //n1代表当前位的值
n2 = (s1[i] - '0' + s2[i] - '0' + n2) / 10; //n2代表进位
res = char(n1 + '0') + res;
}
/*上述循环不能处理第0位的原因在于i的类型是string::size_type，它是非负类型*/
//对于第0位单独处理
n1 = (s1[0] - '0' + s2[0] - '0' + n2) % 10;
n2 = (s1[0] - '0' + s2[0] - '0' + n2) / 10;
res = char(n1 + '0') + res;
if (n2 == 1)
res = "1" + res;
return res;
}
/*
大整数减法
本质上仅仅处理：两整数相减，而且是一大减一小："1234" - "234"
其他情况需转化
1. 小正减大正 => "234" - "1234" = -("1234" - "234")
2. 正减负 => "1234" - "-234" = "1234" + "234"
3. 负减正 => "-1234" - "234" = -("1234" + "234")
4. 负减负 => "-1234" - "-234" = "234" - "1234" = -("1234" - "234")
*/
string sub_int(string s1, string s2)
{
if (s2 == "0")
return s1;
if (s1 == "0")
{
if (s2[0] == '-')
return s2.erase(0, 1);
return "-" + s2;
}
if (s1[0] == '-')
{
if (s2[0] == '-')
{
return sub_int(s2.erase(0, 1), s1.erase(0, 1)); //情况四
}
return "-" + add_int(s1.erase(0, 1), s2); //情况三
}
if (s2[0] == '-')
return add_int(s1, s2.erase(0, 1)); //情况二
//调整s1与s2的长度
string::size_type i, size1, size2;
size1 = s1.size();
size2 = s2.size();
if (size1 < size2)
{
for (i = 0; i < size2 - size1; i++) //在s1左边补零
s1 = "0" + s1;
}
else
{
for (i = 0; i < size1 - size2; i++) //在s2左边补零
s2 = "0" + s2;
}
int t = s1.compare(s2);
if (t < 0) //s1与s2的size同样。但 s1 < s2
return "-" + sub_int(s2, s1);
if (t == 0)
return "0";
//处理本质情况：s1 > s2
string res;
string::size_type j;
for (i = s1.size() - 1; i != 0; i--)
{
if (s1[i] < s2[i]) //不足，需向前借一位
{
j = 1;
while (s1[i - j] == '0')
{
s1[i - j] = '9';
j++;
}
s1[i - j] -= 1;
res = char(s1[i] + ':' - s2[i]) + res;
}
else
{
res = char(s1[i] - s2[i] + '0') + res;
}
}
res = char(s1[0] - s2[0] + '0') + res;
//去掉前导零
res.erase(0, res.find_first_not_of('0'));
return res;
}
string mul_int(string s1, string s2)
{
if (s1 == "0" || s2 == "0")
return "0";
//sign是符号位
int sign = 1;
if (s1[0] == '-')
{
sign *= -1;
s1.erase(0, 1);
}
if (s2[0] == '-')
{
sign *= -1;
s2.erase(0, 1);
}
string::size_type size1, size2;
string res, temp;
size1 = s1.size();
size2 = s2.size();
//让s1的长度最长
if (size1 < size2)
{
temp = s1;
s1 = s2;
s2 = temp;
size1 = s1.size();
size2 = s2.size();
}
int i, j, n1, n2, n3, t;
for (i = size2 - 1; i >= 0; i--)
{
temp = "";
n1 = n2 = n3 = 0;
for (j = 0; j < size2 - 1 - i; j++) temp = "0" + temp;
n3 = s2[i] - '0';
for (j = size1 - 1; j >= 0; j--)
{
t = (n3*(s1[j] - '0') + n2);
n1 = t % 10; //n1记录当前位置的值
n2 = t / 10; //n2记录进位的值
temp = char(n1 + '0') + temp;
}
if (n2)
temp = char(n2 + '0') + temp;
res = add_int(res, temp);
}
if (sign == -1)
return "-" + res;
return res;
}
string divide_int(string s1, string s2, int flag) //flag=1,返回商;flag=0,返回余数
{
string quotient, residue;
if (s2 == "0")
{
quotient = residue = "error";
if (flag == 1)
return quotient;
else
return residue;
}
if (s1 == "0")
{
quotient = residue = "0";
if (flag == 1)
return quotient;
else
return residue;
}
//sign1是商的符号，sign2是余数的符号
int sign1, sign2;
sign1 = sign2 = 1;
if (s1[0] == '-')
{
sign1 *= -1;
sign2 = -1;
s1.erase(0, 1);
}
if (s2[0] == '-')
{
sign1 *= -1;
s2.erase(0, 1);
}
if (compare(s1, s2) < 0)
{
quotient = "0";
residue = s1;
}
else if (compare(s1, s2) == 0)
{
quotient = "1";
residue = "0";
}
else
{
string temp;
string::size_type size1, size2;
size1 = s1.size();
size2 = s2.size();
int i;
if (size2 > 1) temp.append(s1, 0, size2 - 1);
for (i = size2 - 1; i < size1; i++)
{
temp = temp + s1[i];
//试商
for (char c = '9'; c >= '0' ; c--)
{
string t = mul_int(s2, string(1, c));
string s = sub_int(temp, t);
if (s == "0" || s[0] != '-')
{
temp = s;
quotient = quotient + c;
break;
}
}
}
residue = temp;
}
//去除前导零
quotient.erase(0, quotient.find_first_not_of('0'));
residue.erase(0, residue.find_first_not_of('0'));
if (sign1 == -1)
{
quotient = "-" + quotient;
}
if (sign2 == -1)
{
if (residue.empty())
residue = "0";
else
residue = "-" + residue;
}
if (flag == 1) return quotient;
else return residue;
}
string div_int(string s1, string s2)
{
return divide_int(s1, s2, 1);
}
string mod_int(string s1, string s2)
{
return divide_int(s1, s2, 0);
}
int main(void)
{
string s1, s2;
char op;
while (cin >> s1 >> op >> s2)
{
switch (op)
{
case '+':cout << add_int(s1, s2) << endl; break;
case '-':cout << sub_int(s1, s2) << endl; break;
case '*':cout << mul_int(s1, s2) << endl; break;
case '/':cout << div_int(s1, s2) << endl; break;
case '%':cout << mod_int(s1, s2) << endl; break;
default:
cout << "The operator is error!" << endl; break;
}
}
return 0;
}c语言实现大整数加法的实例
#include <stdio.h>
#include <string.h>
#define MAXLEN 1000
char a1[MAXLEN];
char a2[MAXLEN];
static int v1[MAXLEN];
static int v2[MAXLEN];
static int v3[MAXLEN];
int i, j, n, L, z;
void main(void) {
scanf("%d", &n);
for (j = 0; j < n; j++) {
scanf("%s%s", a1, a2);
L = strlen(a1);
for (i = 0; i < L; i++) v1[i] = a1[L - 1 - i] - '0'; //下标越小。位数越高
L = strlen(a2);
for (i = 0; i < L; i++) v2[i] = a2[L - 1 - i] - '0';
for (i = 0; i < MAXLEN; i++) v3[i] = v1[i] + v2[i];
for (i = 0; i < MAXLEN; i++) {
if (v3[i] >= 10) {
v3[i + 1] += 1;
v3[i] = v3[i] % 10;
}
}
printf("Case %d:
", j + 1);
printf("%s + %s = ", a1, a2);
z = 0;
for (i = MAXLEN - 1; i >= 0; i--) {
if (z == 0) {
if (v3[i] != 0) {
printf("%d", v3[i]);
z = 1;
}
}
else {
printf("%d", v3[i]);
}
}
if (z == 0) printf("0");
printf("
");
}
getchar();
getchar();
}
//Sample Input
//3
//0 0
//1 2
//112233445566778899 998877665544332211
//
//Sample Output
//Case 1:
//0 + 0 = 0
//Case 2:
//1 + 2 = 3
//Case 3:
//112233445566778899 + 998877665544332211 = 1111111111111111110 代码下载
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