1002 大数相加

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。

#include <iostream>
#include <string>
using namespace std;
int main()
{
    int n;
    while(cin>>n)//n为case数 
 {
    for(int i=1;i<=n;i++)
    {
        string a,b,c;//3个字符串 
        cin>>a>>b;
        int la=a.length()-1,lb=b.length()-1,jw=0,ta,tb,tt,f=0;
        char tc;
        while(la>=0||lb>=0)
        {
        
            if(la<0) ta=0;
                else ta=a[la]-'0';
            if(lb<0) tb=0;
                else tb=b[lb]-'0';
            tt=jw+ta+tb;//tt为a和b两位相加结果 
            jw=tt/10;//jw为进位
            tc=tt%10+'0';//tc为赋值给字符串c之前的一个中转
            if(tc!='0') f=1;//f为进位标志
            c+=tc;
            la--;lb--;
        }
        if(jw>0)
        {
            f=1;
            tc=jw+'0';
            c+=tc;
        }

        if(i!=1) cout<<endl;
        cout<<"Case "<<i<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        if(f==1)
        {
            for(int j=c.length()-1;j>=0;j--)
                cout<<c[j];
            cout<<endl;
        }
        else  cout<<0<<endl;
    }
    }
    return 0;
}