hdu1002 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): 265006    Accepted Submission(s): 51289

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

 

 

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第一次写大数问题，其实算法还是很好理解的，模拟小学的加法，不过需要将其使用代码实现。

 

题意：a+b求和，不过a，b的位数最大可达1000位，需要使用字符串输入计算，需要注意输出格式，最后一行不空行。

 

附上代码：

#include <iostream>
using namespace std;
#include <string.h>
int main()
{
    int n,i,j=1,p,n1,n2;
    char a[1000],b[1000],c[1001];
    cin>>n;
    while(n--)
    {
        p=0;
        cin>>a>>b;
        cout<<"Case "<<j<<":"<<endl;
        cout<<a<<" + "<<b<<" = ";
        n1=strlen(a)-1;
        n2=strlen(b)-1;
        for(i=0; n1>=0||n2>=0; i++,n1--,n2--)
        {
            if(n1>=0&&n2>=0) c[i]=a[n1]+b[n2]-'0'+p;
            if(n1>=0&&n2<0) c[i]=a[n1]+p;
            if(n2>=0&&n1<0) c[i]=b[n2]+p;
            p=0;
            if(c[i]>'9')
            {
                c[i]-=10;
                p=1;
            }
        }
        if(p==1) cout<<1;
        while(i--)
            cout<<c[i];
        j++;
        if(n!=0)
            cout<<endl<<endl;
        else
            cout<<endl;
    }
    return 0;
}