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): 265243    Accepted Submission(s): 51347

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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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int main()
{
    char a[1005];
    char b[1005];
    int sum[1005];
    int t,lena,lenb;
    scanf("%d",&t);
    int cout=0;
    while(t--)
    {
        cout++;
        scanf("%s",a);
        scanf("%s",b);
        lena=strlen(a);
        lenb=strlen(b);
        memset(sum,0,sizeof(sum));
        int i,j,k,l,f;
        for(i=lena-1,j=lenb-1,k=0; i>=0&&j>=0; i--,j--)
        {
            sum[k+1]=(sum[k]+a[i]-'0'+b[j]-'0')/10;
            sum[k]=(sum[k]+a[i]-'0'+b[j]-'0')%10;
            k++;
            if(i==0&&j!=0)
            {
                for(l=j-1; l>=0; l--)
                {
                    sum[k+1]=(sum[k]+b[l]-'0')/10;
                    sum[k]=(sum[k]+b[l]-'0')%10;
                    k++;
                }
            }
            else if(i!=0&&j==0)
            {
                for(l=i-1; l>=0; l--)
                {
                    sum[k+1]=(sum[k]+a[l]-'0')/10;
                    sum[k]=(sum[k]+a[l]-'0')%10;

                    k++;
                }
            }
        }
        if(sum[k]!=0)//必须放在循环外面
            k++;
        printf("Case %d:
",cout);
        printf("%s + %s = ",a,b);
        for(f=k-1; f>=0; f--)
        {
            {
                printf("%d",sum[f]);
            }
        }
        printf("
");
        if(t!=0)
            printf("
");
    }
}