hdu 1002 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

思路：将数字放入字符数组中，逆序相加

代码实现及讲解：

#include<stdio.h>
#include <iostream>
#include <algorithm>
#include<string.h>
#include<math.h>
using namespace std;
int main()
{
    int T;
    scanf("%d",&T);
    int count = T;
    while (T--)
    {
        char a[1001];	//字符数组记录数字 A ，B ，答案 ans
        char b[1001];
        char ans[1002];
        scanf("%s",a);
        scanf("%s",b);
        strrev(a);		//调用内置函数对字符数组进行逆序操作
        strrev(b);
        int len1 = strlen(a);
        int len2 = strlen(b);
        int i;
        int one = 0;
        int temp;
        if (len1 > len2)
        {
            for (i=0;i<len2;i++)
            {
                temp = a[i] - '0' + b[i] - '0' + one;
                one = temp / 10;
                temp = temp % 10;
                ans[i] = temp + '0';
                }
            while (i<len1)	//将多余的部门直接移入ans 数组
            {
                temp = a[i] - '0' + one;
                one = one / 10;
                temp = temp % 10;
                ans[i] = temp + '0';
                i++;
           }
            if (one)		//判断最后一位是否进位
            {
                ans[i] = one + '0';
                i++;
            }
           ans[i] = '\0';
           strrev(ans);	//ans 数组逆序即为答案
        }
        else
        {
            for (i=0;i<len1;i++)
            {
                temp = a[i] - '0' + b[i] - '0' + one;
                one = temp / 10;
                temp = temp % 10;
                ans[i] = temp + '0';
            }
            while (i<len2)
            {
                temp = b[i] - '0' + one;
                one = one / 10;
                temp = temp % 10;
                ans[i] = temp + '0';
                i++;
            }
            if (one)
            {
                ans[i] = one + '0';
                i++;
            }
            ans[i] = '\0';
            strrev(ans);
        }
        strrev(a);
        strrev(b);
        cout<<"Case "<<count-T<<":"<<endl<<a<<"+"<<b<<"="<<ans<<endl;
        if (T != 0)
        cout<<endl;
    }
    return 0;
}