[ACM]hdu 1002 A + B Problem II (复习大数相加）

A + B Problem II

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 13   Accepted Submission(s) : 4

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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

Author

Ignatius.L 

心得：

通过重写发现数组名真的很重要，不要起那些容易相混的数组名，很容易出错。得到了教训。这类的题只要沉得住气，肯定能做出来的。还有就是细心。思路就不多说了，直接上代码。

代码：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
char a[1004],b[1004];
int  an[1004],bn[1004];
int lena,lenb;

int main()
{
    int t;
    cin>>t;
    for(int k=1;k<=t;k++)
    {
        memset(an,0,sizeof(an));
        memset(bn,0,sizeof(bn));
        cin>>a>>b;
        int c=0,d=0;
        lena=strlen(a);lenb=strlen(b);
        for(int i=lena-1;i>=0;i--)
             an[c++]=a[i]-'0';
        for(int i=lenb-1;i>=0;i--)
             bn[d++]=b[i]-'0';
        int maxlen; maxlen=lena>=lenb?lena:lenb;
        for(int i=0;i<maxlen;i++)
        {
            an[i]+=bn[i];
            if(an[i]>=10)
            {
                an[i+1]++;
                an[i]-=10;
            }
        }
        int j;
        for(j=1003;j>=0;j--)
            if(an[j]!=0)
            break;
        cout<<"Case "<<k<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        for(;j>=0;j--)
            cout<<an[j];
        cout<<endl;
        if(k!=t)
            cout<<endl;
    }
    return 0;
}