A - A + B Problem II

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

#include<iostream>
#include <string>
using namespace std;
int main() {
    
	int T;							//行数
	cin >> T;
	int A[1001], B[1001],C[1001];	//字符串转 int 类型数组
	string a, b;					//字符串
	int al, bl;						//字符串长度


	for (int i = 1; i <= T; i++) {
    	//进行T次
		for (int j = 1; j <= 1001; j++) {
    	//初始化数组全为 0
			A[j - 1] = 0;
			B[j - 1] = 0;
			C[j - 1] = 0;
		}					
		cin >> a >> b;		//输入字符串 a，b
		al = a.length(); bl = b.length();
		int cl = al > bl ? al : bl;		//a,b 中更长的字符串长度

		for (int j = 0; j < al; j++) {
    	//为A数组赋值（逆序）
			A[j] = a[al-j-1] - '0';
		}
		for (int j = 0; j < bl; j++) {
    	//为B数组赋值（逆序）
			B[j] = b[bl-j-1] - '0';
		}
		for (int j = 0; j < cl; j++) {
    	//求和（逆序）
			C[j] = C[j] + A[j] + B[j];
			if (C[j] >= 10) {
    
				C[j] = C[j] - 10;
				C[j + 1] = 1;
			}
		}

		cout << "Case " << i << ":" << endl;
		cout << a << " + " << b << " = ";
		
		if (C[cl] != 0) {
    			//判断最高位是否进位
			cl = cl + 1;
		}
		for (int j = 0; j < cl; j++) {
    	//逆序输出
			cout << C[cl-j-1];
		}
		cout << endl ;
		if (i < T) {
    
			cout << endl;
		}
	}

	return 0;
}