杭电_ACM_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

#include <stdio.h>
#include <string>
#define MAX 1005
int main()
{
    int T, a[MAX], b[MAX], c[MAX], i, j, k, carry;
    char sa[MAX], sb[MAX];
    scanf("%d", &T);
    for (j = 1; j <= T; j++)
    {
        scanf("%s %s", sa, sb);
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        carry = 0;
        //a[0] is the length of a, it's more convenient
        a[0] = strlen(sa);
        b[0] = strlen(sb);
        //translate the char to number and inverse
        for (i = 1; i <= a[0]; i++)
        {
            a[i] = sa[a[0] - i] - '0';
        }
        for (i = 1; i <= b[0]; i++)
        {
            b[i] = sb[b[0] - i] - '0';
        }
        //get the longer one for result
        c[0] = a[0] > b[0] ? a[0] : b[0];
        //add one by one and note the carry
        for (i = 1; i <= c[0]; i++)
        {
            c[i] = (a[i] + b[i] + carry) % 10;
            carry = (a[i] + b[i] + carry) / 10;
        }
        //if the last carry is not zero, you must add the one.
        if (carry != 0)
        {
            c[0]++;
            c[c[0]] = carry; 
        }
        //formatted printing
        printf("Case %d:\n", j);
        for (k = a[0]; k >= 1; k--)
            printf("%d", a[k]);
        printf(" + ");
        for (k = b[0]; k >= 1; k--)
            printf("%d", b[k]);
        printf(" = ");
        for (k = c[0]; k >= 1; k--)
            printf("%d", c[k]);
        printf("\n");
        if (j != T)
            printf("\n");
    }
    return 0;
}

firstly, making your mind clearly.

secondly, in formatted printting, you should care about two lines or one line.