A + B Problem II

A + B Problem II

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 48   Accepted Submission(s) : 19

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

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

Deal(int num)
{
    int sign;
    if(num>=10&&num<20) sign=1;
    else if(num>=20)sign=2;
    else sign=0;
    return sign;
}

int main()
{
    int T,t,LenA,LenB,i,j,LenC,sign;
    char a[1010],b[1010],c[1000];
    scanf("%d",&T);
    t=1;
    while(T--)
    {
       scanf("%s%s",a,b);
       LenA=strlen(a);LenB=strlen(b);
       for(i=LenA-1,j=LenB-1,LenC=0,sign=0;;i--,j--)
       {
            if(i==-1||j==-1)break;
            c[LenC++]=((a[i]-'0')+(b[j]-'0')+sign)%10;
            sign=Deal((a[i]-'0')+(b[j]-'0')+sign);
       }
       if(i==-1)
           for(j;;j--){if(j==-1)break;c[LenC++]=((b[j]-'0')+sign)%10;sign=Deal((b[j]-'0')+sign);}
        else
           for(i;;i--){if(i==-1)break;c[LenC++]=((a[i]-'0')+sign)%10;sign=Deal((a[i]-'0')+sign);}
        if(sign!=0)c[LenC++]=sign;
        printf("Case %d:
",t++);
        printf("%s + %s = ",a,b);
        for(LenC-=1;LenC>=0;LenC--)
            printf("%d",c[LenC]);
        putchar('
');
        if(T!=0)
            putchar('
');
    }
}