大整数加法-优快云博客

本文介绍了一种处理大数加法的算法实现方法，利用字符串形式存储大数，并通过逐位相加的方式完成加法运算。适用于解决超出常规整型变量范围的大数运算问题。

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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

首先要解决的就是存储多位整数的问题.显然,任何C/C++固有类型的变量都无法保存它.最直观的想法是可以用一个字符串来保存它.字符串本质上就是一个字符数组.然后依次从低位到高位进行运算,相加结果大于9即进位.
完整代码：

/******************************************
/*    杭电ACM1002：A + B Problem II
/*    2010/10/1
/*    VC 6.0  by 52coder
/*    persever2009@gmail.com
******************************************/
#include <iostream>
using namespace std;
 
string Sum_calc(string s1,string s2)
{
    int length;
    //比较s1,s2的长度
    if (s1.length()>=s2.length())
    {
        length=s1.length();
        string tmp(s1.length()-s2.length(),'0');
        s2=tmp+s2;
    }
    else
    {
        length=s2.length();
        string tmp(s2.length()-s1.length(),'0');
        s1=tmp+s1;
    }
    string sum(length,'0');
    //carry 进位标志位
    int carry = 0;
    for (int k=length-1;k!=-1;--k)
    {
        //字符'0'对应ASCII码为48
        sum[k]=(s1[k]+s2[k]+carry-96)%10+48;
        carry=(s1[k]+s2[k]+carry-96>9);
    }
    //最高位进位后,进位处理
    if (carry)
        sum='1'+sum;
 
    return sum;
 
}
 
void main()
{
    int T;    //The number of test cases;
    cin>>T;
   //use the dynamic array when not knowing the dimension before compiling;
    string *A = new string[T];
    string *B = new string[T];
    string *Sum = new string[T];
    for (int i=0; i!=T; ++i)
    {
        cin>>A[i]>>B[i];
        Sum[i] = Sum_calc(A[i], B[i]);
    }
    for (int j=0; j!=T; ++j)
   {
    cout<<"Case "<<j+1<<":"<<endl<<A[j]<<" + "<<B[j]<<" = "<<Sum[j];
    if(i!=T-1)
    cout<<endl;
   }