[C++知识库]018:别叫，这个大整数已经很简化了

北大mooc，程序设计与算法（三）

题目

018:别叫，这个大整数已经很简化了!

程序填空，输出指定结果

#include <iostream> 
#include <cstring> 
#include <cstdlib> 
#include <cstdio> 
using namespace std;
const int MAX = 110; 
class CHugeInt {
// 在此处补充你的代码
};
int  main() 
{ 
	char s[210];
	int n;

	while (cin >> s >> n) {
		CHugeInt a(s);
		CHugeInt b(n);

		cout << a + b << endl;
		cout << n + a << endl;
		cout << a + n << endl;
		b += n;
		cout  << ++ b << endl;
		cout << b++ << endl;
		cout << b << endl;
	}
	return 0;
}

输入
多组数据，每组数据是两个非负整数s和 n。s最多可能200位， n用int能表示
输出
对每组数据，输出6行，内容分别是：

样例输入
99999999999999999999999999888888888888888812345678901234567789 12
6 6
样例输出
99999999999999999999999999888888888888888812345678901234567801
99999999999999999999999999888888888888888812345678901234567801
99999999999999999999999999888888888888888812345678901234567801
25
25
26
12
12
12
13
13
14

自己代码

大数加法，重载运算符

#include <iostream> 
#include <cstring> 
#include <cstdlib> 
#include <cstdio> 
using namespace std;
const int MAX = 110; 
class CHugeInt {
	private:
		char s[210];
		int v;
		
	public:
		CHugeInt(char c[]):v(0)
		{
			strcpy(s,c);
		}
		
		CHugeInt(int v_):v(v_)
		{
			memset(s, 0, sizeof(s));
		}
		
		char * Hugejia(char * a, char * b)	//大数相加
		{
			int c1[210];
			int c2[210];
			memset(c1, 0, sizeof(c1));
			memset(c2, 0, sizeof(c2));
			int len_a = strlen(a);
			int len_b = strlen(b);
			int max_l = len_a;
			if(len_b > len_a)
				max_l = len_b;
			
			for(int i = 0; i < len_a; ++i)	//倒过来相加
				c1[i] = a[len_a-i-1] - '0';
			for(int i = 0; i < len_b; ++i)
				c2[i] = b[len_b-i-1] - '0';
			
			for(int i = 0; i <= max_l; ++i)
			{
				c1[i] += c2[i];
				if(c1[i] > 9)	//进位
				{
					c1[i] -= 10;
					++c1[i+1];
				}
			}
			
			if(c1[max_l] != 0)
				++max_l;
			
			for(int i = 0; i <max_l; ++i)
				b[i] = c1[max_l-i-1] + '0';	//再倒回来
			b[max_l] = '\0';	
			return b;
		}
		
		char * operator+ (CHugeInt & d)	//a + b
		{
			static char c[210];	//因为需要返回字符串，不加static函数结束之后字符串就没了
			memset(c, 0, sizeof(c));
			sprintf(c, "%d", d.v);
			return this -> Hugejia(s, c);
		}
		
		friend char * operator+ (int n, CHugeInt & d)	//n + a
		{
			static char c[210];
			memset(c, 0, sizeof(c));
			sprintf(c, "%d", n);
			return d.Hugejia(d.s, c);
		}
		
		char * operator+ (int n)	//a + n
		{
			static char c[210];
			memset(c, 0, sizeof(c));
			sprintf(c, "%d", n);
			return this -> Hugejia(s, c);
		}
		
		void operator+= (int & n)	//b += n
		{
			v = v + n;
		}
		
		int & operator++ ()	//前自增
		{
			v = v + 1;
			return v;
		}
		
		int operator++ (int k)	//后自增
		{
			int temp = v;
			v = v + 1;
			return temp;
		}
		
		friend ostream & operator<< (ostream & os, const CHugeInt & d)	//流插入 <<
		{
			os << d.v;
			return os;
		}
};

int  main() 
{ 
	char s[210];
	int n;

	while (cin >> s >> n) {
		CHugeInt a(s);
		CHugeInt b(n);

		cout << a + b << endl;
		cout << n + a << endl;
		cout << a + n << endl;
		b += n;
		cout << ++b << endl;
		cout << b++ << endl;
		cout << b << endl;
	}
	return 0;
}