C++：HDU1002 A + B Problem II（大数加法）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 409341    Accepted Submission(s): 79315

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

 

Author

Ignatius.L

解题思路：这是道大数加法问题，由于输入的数字可能会比较大，所以我们把输入当做字符串来存；然后分别倒叙存入数组中，让2个数组进行计算，从第一项开始，像小学加法算术一样，大于10就前一位进1，存入一个新数组；最后把新数组倒序输出。

例：1234+567

	0	1	2	3
S1	1	2	3	4
S2	5	6	7

	0	1	2	3
S3	4	3	2	1
S4	7	6	5
s	1	0	8	1

然后把s倒序输出

1234 + 567 = 1801

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    cin>>n;
    for(int k=1;k<=n;k++)
    {
        char s1[1005],s2[1005];
        int s3[1005],s4[1005],s[1100]={0};
        scanf("%s%s",s1,s2);
        int l1=strlen(s1);
        int l2=strlen(s2);
        for(int i=l1-1,j=0;i>=0;i--,j++)//倒序存入数组，记得-1，因为字符串是从0开始存入的，最后一位是s3[l1-1]
            s3[j]=s1[i]-'0';
        for(int i=l2-1,j=0;i>=0;i--,j++)
            s4[j]=s2[i]-'0';
        for(int i=0;i<min(l1,l2);i++)//先将min(l1,l2)存入数组
        {
            s[i]=s3[i]+s4[i]+s[i];//加s[i]是因为后一位可能会进1
            if(s[i]>=10)
            {
                s[i+1]=1;//如果和大于10，就要前一位进1
                s[i]=s[i]-10;
            }
        }
        for(int i=min(l1,l2);i<max(l1,l2);i++)//把剩下的存入数组
        {
            if(l1>l2)//判断是哪个剩下了
                s[i]=s3[i]+s[i];
            if(l1<l2)
                s[i]=s4[i]+s[i];
        }
        cout<<"Case "<<k<<':'<<endl;//注意输出格式
        cout<<s1<<" "<<'+'<<" "<<s2<<" "<<'='<<" ";
        for(int i=max(l1,l2)-1;i>=0;i--)//倒序输出
            cout<<s[i];
        if(k!=n)//注意格式，除了最后一组测试是一个换行，其他都是两个换行
            cout<<endl<<endl;
        else
            cout<<endl;
    }
}