杭电1002（大数A+B）之 A + B Problem II

最新推荐文章于 2021-01-30 13:04:09 发布

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本文介绍了一个简单的编程挑战：给定两个大整数A和B，计算它们的和。输入包含多个测试案例，每个案例由两个大整数组成。文章提供了一段使用C++实现的AC代码，该代码通过字符串操作解决了大整数相加的问题。

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


AC代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

string a,b;

int main()
{
    int t;
    char s;
    cin>>t;
    for(int l=1;l<=t;l++)
    {
        cin>>a>>b;
        s='0';
        string c=a;
        string d=b;
        int n=a.length();
        int m=b.length();
        if(n>=m)
        {
            for(int i=n-1,j=m-1;i>=0 || j>=0;i--)
            {
                if(i>=0 && j>=0)
                {
                    if(a[i]+b[j]-'0'-'0'>=10)
                    {
                        if(i-1>=0)
                            a[i-1]++;
                        else
                            s='1';
                        a[i]=(a[i]+b[j]-'0'-10);
                    }
                    else
                    {
                        a[i]=(a[i]+b[j]-'0');
                    }
                }
                if(j>=0) j--;
            }
        }
        else
        {
            for(int i=m-1,j=n-1;i>=0 || j>=0;i--)
            {
                if(i>=0 && j>=0)
                {
                    if(a[j]+b[i]-'0'-'0'>=10)
                    {
                        if(i-1>=0)
                            b[i-1]++;
                        else
                            s='1';
                        b[i]=(a[j]+b[i]-'0'-10);
                    }
                    else
                    {
                        b[i]=(a[j]+b[i]-'0');
                    }
                }
                if(j>=0) j--;
            }
        }
        printf("Case %d:\n",l);
        cout<<c<<" + "<<d<<" = ";
        if(s=='1') cout<<s;
        if(n>=m) cout<<a<<endl;
        else cout<<b<<endl;
        if(l<t) cout<<endl;
    }
    return 0;
}