A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 72497 Accepted Submission(s): 13352

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

巧妙的解題方法，將字符型數(shù)據(jù)轉(zhuǎn)換為一位的整型數(shù)據(jù)加進位少欺。

#include<stdio.h>
#include <stdlib.h>
#include <string.h>
void main()
{
    int i, j, k, l, n, m;
    char a[1010],b[1010];
    int h[1100] ;
    scanf("%d", &n);
    for (i = 0;i < n;i++)
    {
        for (j=0;j<1010;j++)
            h[j]=0;
        j=0;
        printf("Case %d:\n",i+1);
        scanf("%s%s", a, b);
        printf("%s+%s=",a,b);
        k = strlen(a) - 1;
        l = strlen(b) - 1;
        while (k >= 0 && l >= 0)
        {
            if (h[j] + (a[k] - '0') + (b[l] - '0') >= 10)
            {
                h[j] = (h[j] + (a[k] - '0') + (b[l] - '0')) % 10;
                h[j + 1]++;
                
            }
            else
                h[j] = h[j] + (a[k] - '0') + (b[l] - '0');
            j++;
            k--;
            l--;
        }
            while(k >= 0)
        {
            if (h[j] + (a[k] - '0') >= 10)
            {
                h[j] = (h[j] + a[k] - '0') % 10;
                h[j + 1]++;
            }
            else
                h[j] = h[j] + a[k] - '0';
            j++;
            k--;
        }
        while(l >= 0)
        {
            if (h[j] + (b[l] - '0') >= 10)
            {
                h[j] = (h[j] + b[l] - '0') % 10;
                h[j + 1]++;
            }
            else
                h[j] = h[j] + b[l] - '0';
            j++;
            l--;
        }
        for (;h[j] == 0;j--);
        for (k = j;k >= 0;k--)
            printf("%d", h[k]);
        printf("\n");
        if(i!=n-1)
            printf("\n");
    }
}