A + B Problem II（1002）

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

 

思路：字符串输入，将两个字符串倒叙输入两个新整型数组（一开始一直在纠结如何倒叙相加）。然后相加进位。注意最后要判断进位是否为0 .

 

注意！！！一定要将两个新的整型数组初始化为0（这长度不一的两个数倒叙相加时不会出错），另外最好用一个新的数组来接受相加结果。。

 

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

int main()
{
    int n , ans = ;
    cin >> n ;
    while(n --)
    {
        ans++ ;
        char a[] = {},  b[] ={} ;
        int c[] ={}, d[] = {} , e[] ={};
        scanf("%s%s" , &a , &b);
        int len1 = strlen(a) ;
        int len2 = strlen(b) ;
        int len = max(len1 , len2);
        int j =  ;
        for(int i = len1 -  ; i >=  ; i--)
        {
            c[j++] = a[i] -  ;
        }
        j =  ;
        for(int i = len2 -  ; i >=  ; i--)
        {
            d[j++] = b[i] -  ;
        }
        int k =  ;
        for(int i =  ; i < len ; i++)
        {
            e[i] = (c[i] + d[i] + k) %  ;
            k =  (c[i] + d[i] + k) /  ;
        }
        cout << "Case " << ans << ':' << endl ;
        printf("%s + %s = " , a , b);
        if(!k)
        {
            for(int i = len -  ; i >=  ; i--)
                cout << e[i] ;
            cout << endl ;
        }
        else
        {
            e[len] = k ;
            for(int i = len ; i >=  ; i--)
                cout << e[i] ;
            cout << endl ;
        }
        if(n != )
            printf("\n");

    }
    return ;
}