HDU 1002 A + B Problem II （大数加法） A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 487411 Accepted Submission(s): 94074

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

题目大意与分析：

就是求两个大数的加法，两个用例之间还要有空格，发在这里记一下模板

#include<bits/stdc++.h>

using namespace std;
int n,lena,lenb,ai,bi,jin,lenmax,p,q,i,t=0;
int anss[100000];
int main()
{
    cin>>n;
    while(n--)
    {
        t++;
        memset(anss,0,sizeof(anss));
        jin=0;
        string a;
        string b;
        cin>>a;
        cin>>b;
        lena=a.size();
        lenb=b.size();
        lenmax=max(lena,lenb);
        p=lena-1;
        q=lenb-1;
        for(i=lenmax-1;i>=0;i--)
        {
            if(p<0)
            ai='0';
            else
            ai=a[p];
            if(q<0)
            bi='0';
            else
            bi=b[q];        
            anss[i]=(ai-'0'+bi-'0'+jin)%10;
            jin=(ai-'0'+bi-'0'+jin)/10;
            p--;
            q--;
        }
        cout<<"Case "<<t<<':'<<endl;
        cout<<a<<" + "<<b<<" = ";
        if(jin)
        cout<<jin;
        for(i=0;i<lenmax;i++)
        cout<<anss[i];
        cout<<endl;
        if(n)
        cout<<endl;
    }
}