hdoj 1002 A+B(2)

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>
#include <math.h>
#define N 1010
int t;
char str1[N], str2[N];
int sum1[N], sum2[N];
int main()
{
    int len1, len2, num;
    scanf("%d", &t);
    //getchar();
    num = 1;
    while(t--)
    {
        memset(sum1, 0, sizeof(sum1));
        memset(sum2, 0, sizeof(sum2)); 
        scanf("%s", str1);
        scanf("%s", str2);
        len1 = strlen(str1);
        len2 = strlen(str2);
        int i, j;
        for(i = len1-1, j = 0; i >= 0; i--)
            sum1[j++] = sum1[j] + str1[i] - '0';
        for(i = len2-1, j = 0; i >= 0; i--)
            sum2[j++] = sum2[j] + str2[i] - '0';
        for(i = 0; i < N; i++)
        
        {
            sum2[i] += sum1[i];
            if(sum2[i] >= 10)
            {
                sum2[i] -= 10;
                sum2[i+1]++;
            }
        }
        for(i = N-1; i>=0 && sum2[i]==0; i--)
            ;
        printf("Case %d:\n", num++);
        printf("%s + %s = ", str1, str2);
        if(i>=0)
        {
            for(; i >= 0; i--)
                printf("%d", sum2[i]);
        }
        else
            printf("0");
        if(t)//注意格式 
            printf("\n\n");
        else
            printf("\n");
    }
    return 0;
}