大数相加

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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): 299941 Accepted Submission(s): 57818

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

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string a,b,c;
    long long int n,i,j,x,y,k,m;
    cin>>n;
    {
        for(m=1;m<=n;m++)
        {
            cin>>a>>b;
            cout<<"Case "<<m<<":"<<endl;
            cout<<a<<" + "<<b<<" = ";
            x=a.size();
            y=b.size();
            k=0;
            if(x>y)
            for(i=0;i<=x;i++)c[i]='0';
            else for(i=0;i<=y;i++)c[i]='0';
            for(i=x-1,j=y-1;;i--,j--)
            {
                if(i>=0&&j>=0)
                {
                    if(a[i]+b[j]-48<='9'){c[k]+=a[i]+b[j]-96;k++;}
                    else {c[k]+=a[i]+b[j]-106;k++;c[k]='1';}
                }
                else if(i<0&&j>=0)
                {
                    c[k]+=b[j]-48;
                    if(c[k]>'9'){c[k]='0';k++;c[k]='1';}
                    else k++;
                }
                else if(j<0&&i>=0)
                {
                    c[k]+=a[i]-48;
                    if(c[k]>'9'){c[k]='0';k++;c[k]='1';}
                    else k++;
                }
                else break;
            }
            if(c[k]!='0')cout<<c[k];
            for(i=k-1;i>=0;i--)
                cout<<c[i];
            cout<<endl;
            if(m<n)cout<<endl;

        }
    }
    return 0;
}

最简单的大数问题之一，没什莫可说的，主要是字符数组处理

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