大家都有过高精度类型的经历，比如让你算 A+B 的答案，正常人肯定就是认为就直接 A+B 了，但是如果 A,B 的范围都超过了 long long 的范围。那么 C++肯定是用long long算不出来的，所以我们要进行字符串的模拟
A+B $(A,B>=0)$
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string add(string a, string b)
{
int hex = 0;
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
int len1 = a.length();
int len2 = b.length();
string ans;
for (int i = 0, j = 0; i < len1 || j < len2; i++, j++)
{
int num1 = 0, num2 = 0;
if (i < len1)
num1 = a[i] - '0';
if (j < len2)
num2 = b[i] - '0';
ans = char((num1 + num2 + hex) % 10 + '0') + ans;
hex = (num1 + num2 + hex) / 10;
}
if (hex)
ans = char(hex + '0') + ans;
return ans;
}
int main()
{
string a, b;
while (cin >> a >> b)
cout << add(a, b) << endl;
}
A*B $(A,B>=0)$
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string mult(string a, string b)
{
string ans;
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
int len1 = a.length(), len2 = b.length(), result[len1 + len2] = {0};
for (int i = 0; i < len1; i++)
for (int j = 0; j < len2; j++)
result[i + j] += (a[i] - '0') * (b[j] - '0');
for (int i = 0; i < len1 + len2 - 1; i++)
{
result[i + 1] += result[i] / 10;
result[i] %= 10;
}
int i = len1 + len2 - 1;
while (result[i] == 0 && i > 0)
i--;
for (; i >= 0; i--)
ans += char(result[i] + '0');
return ans;
}
int main()
{
string a, b;
while (cin >> a >> b)
cout << mult(a, b) << endl;
}
A-B $(A,B>=0)$
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string sub(string a, string b)
{
int len1 = a.length(), len2 = b.length(), result[max(len1, len2)]{0}; //假设a是最大的，b是最小的
bool flag = false;
string ans;
if (len1 < len2)
swap(a, b), swap(len1, len2), flag = true;
else if (len1 == len2)
{
int i;
for (i = 0; i < len1 && a[i] == b[i]; i++);
if (a[i] < b[i])
swap(a, b), flag = true;
}
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
int i;
for (i = 0; i < len2; i++)
if (a[i] >= b[i])
result[i] = a[i] - b[i];
else
result[i] = a[i] + 10 - b[i], a[i + 1] -= 1;
while (i < len1)
result[i] = a[i] - '0', i++;
for (i = len1 - 1; i > 0; i--)
if (result[i] == 0)
len1--;
else
break;
for (i = 0; i < len1; i++)
ans = char(result[i] + '0') + ans;
if (flag)
ans = "-" + ans;
return ans;
}
int main()
{
string a, b;
while (cin >> a >> b)
cout << sub(a, b) << endl;
}
