H - Problem H

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

题目大概意思：输入两个很大很大的数字，意味着这不是简单的整型类型，然后按要求格式输出两个数字的和

我的见解：看起来很简单，很多首先考虑字符串，然后将短的补充至与长的相等，jian用ASCⅡ码进行位与位的运算，再建多一个字符数组装他们的和，要着重考虑递进的问题哦，，，，，想到了一个更好的方法，可以减少字符组的使用，以后有时间再贴出来了，再下面有个更高级的算法，用了高级补充字符函数memset。

代码如下：

#include <iostream>
#include<cstring>
using namespace std;
int main()
{
	int n, j, k, p, h, i = 0;
	cin >> n;
	while (n)
	{
		i++; n--;
		char a[1001], b[1001], c[1002] = { 0 }, d[1001] = { 0 }, l[1001];
		cin >> a >> b;
		j = strlen(a);
		k = strlen(b);
		if (j > k)
		{
			for (int r = j - k; r < j; r++)
			{
				d[r] = b[r - (j - k)];
			}
			strcpy_s(l, a);
			h = j;
		}
		else if (j < k)
		{
			for (int r = k - j; r < k; r++)
			{
				d[r] = a[r - (k - j)];
			}
			strcpy_s(l, b);
			h = k;
		}
		else
		{
			h = j;
			strcpy_s(l, b);
			strcpy_s(d, a);
		}
		for (p = h; p >= 1; p--)
		{
			if (((l[p - 1] + d[p - 1]) + c[p] < 106) && d[p - 1] != \'\\0\')
			{
				c[p] += l[p - 1] + d[p - 1] - 48;
			}
			else if ((l[p - 1] + c[p] < 58) && d[p - 1] == \'\\0\')
			{
				c[p] += l[p - 1];
			}
			else if ((l[p - 1] + c[p] >= 58) && d[p - 1] == \'\\0\')
			{
				c[p] += l[p - 1] - 10;
				c[p - 1] += 1;
			}
			else

			{
				c[p] += l[p - 1] + d[p - 1] - 58;
				c[p - 1] += 1;
			}

		}
		if (c[0] == \'\\0\')
		{
			c[0] = \' \';
			cout << \"Case \" << i << \":\" << endl;
			cout << a << \" + \" << b << \" =\" << c << endl;
		}
		else
		{
			c[0] += 48;
			cout << \"Case \" << i << \":\" << endl;
			cout << a << \" + \" << b << \" = \" << c << endl;
		}
		if (n) cout << endl;


	}
}

高级代码方法：（少部分借鉴网上大神代码）

#include <iostream>
using namespace std;

void main()
{
	char a[1002], b[1002], c[1002], d[1002], e[1002], f[1002];
	int n, Lenmax, t = 0, p = 1;
	cin >> n;
	while (n--)
	{
		memset(a, \'0\', sizeof(a));
		memset(b, \'0\', sizeof(b));
		memset(c, \'0\', sizeof(c));
		memset(d, \'0\', sizeof(d));
		memset(e, \'0\', sizeof(e));
		memset(f, \'\\0\', sizeof(f));
		cin >> a >> b;
		if (strlen(a) > strlen(b))Lenmax = strlen(a);
		else Lenmax = strlen(b);
		for (int i = strlen(a) - 1, x = 0; i >= 0; i--, x++)//将数字反序  99789→98799
		{
			c[x] = a[i];
		}
		for (int i = strlen(b) - 1, x = 0; i >= 0; i--, x++)//将数字反序  789→987
		{
			d[x] = b[i];
		}
		for (int k = 0; k < Lenmax; k++)//将数字相加 
		{
			t += c[k] - \'0\';
			t += d[k] - \'0\';
			e[k] = (t % 10) + \'0\';//9+9=18 取8	9+8+1=18 取8  7+7+1=15 取5  9+0+1=10 取0  9+0+1=10 取0 0+0+1 取1
			if (t >= 10 && k == Lenmax - 1)Lenmax++;
			if (t >= 10)t = 1;	  //18>10进1位	18>10进一位	  15>10 进一位  10=10 进一位  10=10 进一位
			else t = 0;
		}

		for (int z = Lenmax - 1, i = 0; z >= 0; z--, i++)
		{
			f[i] = e[z];
		}
		cout << \"Case \" << p++ << \":\" << endl << a << \" + \" << b << \" = \" << f << endl;
		if (n)cout << endl;
	}
}