[HDOJ#1002]A + B Problem II

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

 

 

My solution:

 

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

int main(void)
{
    int n,i,j;
    char a[1001],b[1001];
    cin>>n;
    void plus(char *a,char *b);
    for (i=1;i<=n;i++)
    {
        cin>>a>>b;
        cout<<"Case "<<i<<":"<<endl<<a<<" + "<<b<<" = ";
        plus(a,b);
        cout<<endl;
        if (i!=n) cout<<endl;
    }
    return 0;
}
void plus(char *a,char *b)
{
    int lena,lenb,k=0,t;
    
    lena=strlen(a);
    lenb=strlen(b);
    int i,j,na[1001],nb[1001],sum[1001]={0};
    for (i=0;i<lena;i++)
        na[i]=a[i]-48;
    for (j=0;j<lenb;j++)
        nb[j]=b[j]-48;
    while(i>=0&&j>=0)
    {
        sum[k]=na[i]+nb[j];
        j--;
        i--;
        k++;
    }
    if (j>=0)
        while (j>=0)
        {
            sum[k]=nb[j];
            j--;
            k++;
        }
    else if (i>=0)
        while(i>=0)
        {
            sum[k]=na[i];
            i--;
            k++;
        }
for (t=1;t<k;t++)
    if (sum[t]>=10)
    {
        sum[t]%=10;
        sum[t+1]++;
    }
while(sum[k]==0)
    k--;
for (t=k;t>0;t--)
    cout<<sum[t];
}