HDU 1002 A + B Problem II

 

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


[align=left]Problem Description[/align]
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
1 2
112233445566778899 998877665544332211

 

[align=left]Sample Output[/align]

Case 1:
1 + 2 = 3

Case 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;

}