HDU__1002A + B Problem II(大数加法)

Problem Description

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


注意事项都在代码里注释了,还是爱犯很多小毛病

 #include
 #include
 #include
 #include
 #include
 using namespace std;
 int main()
 {
 	int t,putcnt = 0;
 	string s1,s2;
 	vector a,b,c;
 	cin>>t;
 	while(t--)
 	{
 		cin>>s1>>s2;
// 		a.clear();
// 		b.clear();
// 		c.clear();//清空后并没有自动赋值为0!! 
 		//逆序 ——并没有实现逆序!!! 
 		for(int i = s1.size() - 1; i >= 0;i--)
 		{
 			a.push_back(s1[i] - '0');
		}
//		for(int i = 0; i < a.size();i++)
//		{
//			cout<= 0;i--)
 		{
 			b.push_back(s2[i] - '0');
		}
//		cout<= a.size())//补位操作 
			{
				a.push_back(0);
			}
			if(i >= b.size())
			{
				b.push_back(0);
			}
			
			push = (a[i] + b[i] + is_jin) % 10;
			c.push_back(push);
			is_jin = (a[i] + b[i] + is_jin) / 10;
			
			
		}
		if(is_jin)
		{
			c.push_back(1);
		}
		// 输出
		printf("Case %d:\n",++putcnt);
		cout<= 1)
		 {
		 	cout<



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