hdu1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 100391    Accepted Submission(s): 19038

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

 

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
char a[1000+10];
char b[1000+10];
int  convert(char *p)
{
    int len=strlen(p);
    int j=len-1;
    int i=0;
    while(i<j)
    {
        char temp=p[i];
        p[i]=p[j];
        p[j]=temp;
        i++;j--;
    }
    return len;
}
void cal(char *p,char *q,int lena,int lenb)
{
    int lenc=max(lenb,lena);
    if(lena<=lenb)
    {
        for(int i=lena;i<=lenb;i++)
        {
            p[i]=q[i];
        }
        for(int i=0;i<lena;i++)
        {
            p[i]+=q[i]-'0';
        }
    }
    else
    {
        for(int i=0;i<lenb;i++)
        {
            p[i]+=q[i]-'0';
        }
    }
    for(int i=0;i<lenc-1;i++)
    {
        p[i+1]+=(p[i]-'0')/10;
        p[i]=(p[i]-'0')%10+'0';
    }
    cout<<int(p[lenc-1]-'0');
    for(int i=lenc-2;i>=0;i--)
    {
        cout<<p[i];
    }
    cout<<endl;
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        memset(a,'0',sizeof(a));
        memset(b,'0',sizeof(b));
        scanf("%s%s",a,b);
        printf("Case %d:\n",i);
        cout<<a<<" + "<<b<<" = ";
        int lena=convert(a);
        int lenb=convert(b);
        cal(a,b,lena,lenb);
        if(i!=n)cout<<endl;
    }
    return 0;
}