在計算過大的數字時，我們可以使用字串進行儲存，再模擬計算過程，結果也用字串儲存，最後輸出這個字串

加法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;

int main()
{
	string a, b;
	cin >> a >> b;
	int l1 = a.size();
	int l2 = b.size();
	if(l1 < l2)
	{
		string temp = a;
		a = b;
		b = temp;
	}
	l1--, l2--;
	int G = 0;
	while(l2 >= 0)
	{
		int S = a[l1] - '0' + b[l2] - '0'+ G;
		G = S/10;
		S = S % 10;
		a[l1] = S + '0';
		l1--, l2--;
	}
	while(l1 >= 0)
	{
		int S = a[l1] - '0' + G;
		G = S/10;
		S = S % 10;
		a[l1] = S + '0';
		l1--, l2--;
	}
	if(G != 0)
		a = '1' + a;
	cout << a << endl;
}
 

上面那個雖然能計算，但是有紕漏，再來一個修改的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;

int main()
{
	string a, b;
	cin >> a >> b;
	cout << a << " + " << b << " = ";
	int l1 = a.size();
	int l2 = b.size();
	if(l1 < l2)
	{
		string temp = a;
		int t = l1;
		a = b;
		l1 = l2;
		b = temp;
		l2 = t;
	}
	l1--, l2--;
	int G = 0;
	while(l2 >= 0)
	{
		int S = a[l1] - '0' + b[l2] - '0'+ G;
		G = S/10;
		S = S % 10;
		a[l1] = S + '0';
		l1--, l2--;
	}
	while(l1 >= 0)
	{
		int S = a[l1] - '0' + G;
		G = S/10;
		S = S % 10;
		a[l1] = S + '0';
		l1--, l2--;
	}+
	if(G != 0)
		a = '1' + a;
	cout << a << endl;
	return 0;
}