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
难点是，数据类型最长有32位（4字节或者2字），数值范围是-2147483648~2147483648或者0~4294967295，但题目中指出输入数据位数长度可以达到1000位，10^999>>4294967295,故不能用常规方法
具体解决方法是，将数字利用字符串的形式表示，每个字符都是数字，1000个连续字符也没问题，再将两个不同字符串相加得到最终结果。
有一次提交时，出现了“Presentation Error”的错误，缘由是输出结果的格式不符合要求，比方少个空格什么的。
1 #include2 #include 3 using namespace std; 4 int main() 5 { 6 int n; 7 while(cin>>n)//n为case数 8 { 9 for(int i=1;i<=n;i++)10 {11 string a,b,c;//3个字符串 12 cin>>a>>b;13 int la=a.length()-1,lb=b.length()-1,jw=0,ta,tb,tt,f=0;14 char tc;15 while(la>=0||lb>=0)16 {17 18 if(la<0) ta=0;19 else ta=a[la]-'0';20 if(lb<0) tb=0;21 else tb=b[lb]-'0';22 tt=jw+ta+tb;//tt为a和b两位相加结果 23 jw=tt/10;//jw为进位24 tc=tt%10+'0';//tc为赋值给字符串c之前的一个中转25 if(tc!='0') f=1;//f为进位标志26 c+=tc;27 la--;lb--;28 }29 if(jw>0)30 {31 f=1;32 tc=jw+'0';33 c+=tc;34 }35 36 if(i!=1) cout<