HDU 1002 A + B Problem II

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 161989 Accepted Submission(s): 30809

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

这里要运用高精度来运算，用数组来存储数据，可以先定义字符数组，然后在进行字符与数字之间的转换，，最主要的是要考虑到进位的问题（关键）。下面是我的代码：

#include<iostream>
#include<cstring>
using namespace std;

const int MAXN=1000+10;
char str1[MAXN];
char str2[MAXN];
int str3[MAXN];
int str4[MAXN];
int str5[MAXN];

int main()
{
    int T;
    cin>>T;
    int k=0;
    while(T--)
    {
        int i=0;
        static int j=1;
        memset(str1, 0, sizeof(str1));
        memset(str2, 0, sizeof(str2));
        memset(str3, 0, sizeof(str3));
        memset(str4, 0, sizeof(str4));
        cin>>str1;
        cin>>str2;
        int len1=strlen(str1);
        for( i=0; i<len1; i++)
        {

            str3[i]=str1[len1-i-1]-'0';
        }
        int len2=strlen(str2);
        for(i=0; i<len2; i++)
        {
            str4[i]=str2[len2-i-1]-'0';
        }
        int c=0;
        for( i=0; i<MAXN; i++)
        {
            str5[i]=str3[i]+str4[i]+c;
            c=str5[i]/10;
            str5[i]%=10;
        }
        for(i=MAXN-1; i>=0 && str5[i]==0;i--);
        cout<<"Case "<<j++<<":"<<endl;
        cout<<str1<<" + "<<str2<<" = ";
        if(i>=0)
        {
            for(; i>=0; i--)
            {
               cout<<str5[i];
            }

        }
        cout<<endl;
        if(T-1>=0)
        {
            cout<<endl;
        }

}
return 0;
}