HDU-1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 287467    Accepted Submission(s): 55255

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<stdio.h>
#include<string.h>
int main()
{
    int n,j;
    scanf("%d",&n);
    for(j=0;j<n;j++)
    {
        int a[1200]={0},b[1200]={0},c[1200]={0},i,long1,long2,k;
        char a1[1200],b1[1200];
        scanf("%s",a1);
        scanf("%s",b1);
        long1=strlen(a1);
        long2=strlen(b1);
        for(i=0;i<long1;i++)
            a[i]=a1[long1-i-1]-'0';
        for(i=0;i<long2;i++)
            b[i]=b1[long2-i-1]-'0';
            
          k=long1<long2?long2:long1;
        for(i=0;i<k;i++)
            c[i]=a[i]+b[i];
        for(i=0;i<k;i++)
            if(c[i]>=10)
            {
                c[i+1]+=c[i]/10;
                c[i]%=10;
            }
            printf("Case %d:
",j+1);
            printf("%s + %s = ",a1,b1);
            if(c[k]==0)
                k--;
            for(i=k;i>=0;i--)
                printf("%d",c[i]);
            if(j==n-1)
                printf("
");
            else
                printf("

");
    }
    return 0;
}