hduoj #1002 A + B Problem II

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002
题目描述如下：

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)

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  

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题目大意： 大数相加
题目思路： 从最低位开始一位一位的加。注意进位，注意清除前缀的0
C++ 代码如下：

#include 
#include 

using namespace std;

int main()
{
    int n;
    char a[1001],b[1001];
    short c[1001];
    cin >> n;
    for (int i=0; i> a >> b;
        alen = strlen(a);
        blen = strlen(b);
        clen = 0;
        maxlen = alen/blen ? alen : blen;
        int ai,bi,s=0;
        for (int j=0; j=0 && bi>=0)
                s = a[ai]+b[bi]-2*'0'+s/10;
            else if (ai>=0)
                s = a[ai]-'0'+s/10;
            else if (bi>=0)
                s = b[bi]-'0'+s/10;
            c[clen++] = s%10;
        }
        if (s/10)
            c[clen++] = s/10;
        for (int k=clen-1; k>=0; k--)
            if (c[k])
                break;
            else
                clen--;
        cout << "Case " << i+1 << ":" << endl;
        cout << a <<" + " << b << " = ";
        for (int k=clen-1; k>=0; k--)
            cout << c[k];
        cout << endl;
        if (n-i-1)
            cout << endl;
    }
    return 0;
}