杭电OJ----1002A + B问题II（超大数计算问题）

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

问题描述

对您来说，我有一个非常简单的问题。给定两个整数A和B，您的工作是计算A + B的总和。

输入项

输入的第一行包含一个整数T（1 <= T <= 20），它表示测试用例的数量。然后是T行，每行由两个正整数A和B组成。请注意，这些整数非常大，这意味着您不应使用32位整数对其进行处理。您可以假设每个整数的长度不超过1000。

输出量

对于每个测试用例，您应该输出两行。第一行是“ Case＃：”，＃表示测试用例的编号。第二行是等式“ A + B = Sum”，Sum表示A + B的结果。请注意，该等式中有一些空格。在两个测试用例之间输出空白行。

样本输入

1 2

112233445566778899 998877665544332211

样本输出

情况1： 1 + 2 = 3

情况2： 112233445566778899 + 998877665544332211 = 1111111111111111110

解答：

#include<cstdio>
#include<iostream>
#include<string>
using namespace std;
int main()
{
   int p,A,B,num=1,sum[1000];
   string a, b;
   cin >> p;
   while (p--)
   {
       int m=0, n=0, k = 0, temp=0;
       cin>>a>>b;
       A = a.length();
       B = b.length();
       A -= 1;
       B -= 1;
       while (A>=0&&B>=0)
       {
       m =a[A]-'0';
       n =b[B]-'0';
       sum[k++] = (m+n+temp)%10;
       temp = (m+n+temp)/10;
       A--;
       B--;
       }
       if (A > B)
       {
           while (A>=0)
           {
               n = a[A] - '0';
               sum[k++] = (n+temp) % 10;
               temp = (n+temp) / 10;
               A--;
           }
       }
       if (A<B)
       {
           while (B>=0)
           {
               n = b[B] - '0';
               sum[k++] = (n + temp) % 10;
               temp = (n + temp) / 10;
               B--;
           }
       }
       sum[k] = temp;
       cout << "Case " << num << ':' << endl;
       cout << a <<" + "<<b << " = ";
       num++;
   for(int i=k-1;i>0;i--)
       {
       cout << sum[i];
       }
   cout << sum[0]<<endl;
   if (p !=0) cout << endl;
   }
   return 0;
}

杭电OJ----1002A + B问题II（超大数计算问题）

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