大数相加

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 x, sa[1001], sb[1001], k = 1, i, j, m, n, t, d;
        char a[1001], b[1001];
        scanf("%d", &x);
        while( x-- ){

                //int sa[1001] = {0};
                //int sb[1001] = {0};#1
                memset(sa, 0, sizeof(sa));
                memset(sb, 0, sizeof(sb));
                scanf("%s", a);
                scanf("%s", b);
                m = strlen(a);
                n = strlen(b);
                for( i = 0, j = m - 1; j >= 0; j-- )
                {

                        //sa[i++] = a[j];
                        sa[i++] = a[j] - '0';

                }
                for( i = 0, j = n - 1; j >= 0; j-- )
                {

                        //sb[i++] = b[j];
                        sb[i++] = b[j] - '0';
                }
                t = m >= n ? m : n;/*找出最长的长度*/
                for( i = 0; i < t; i++ )
                {

                        sa[i] += sb[i];
                        if( sa[i] >= 10 )
                        {

                                sa[i] %= 10;    
                                sa[i + 1]++;

                        }

                }
                //d = (sa[i] == 1) ? t : t - abs(m - n) ;
                d = (sa[i] == 1) ? t : t - 1;/*如果最高位还进1位，那就从它的上位开始输出*/
                printf("Case %d:
%s + %s = ", k++, a, b);
                while( d >= 0 ){

                        //printf("%d", d--);
                          printf("%d", sa[d--]);    

                }
                putchar('
');
                //putchar('
');细节问题：最后一次测试无空行
                //return 0;
                if( x == 0 )
                        continue;
                putchar('
');

        }
        return 0;

}

 

note:

#1只有定义变量才可以全部赋值，但只能是0。一般不喜欢这样做，所以调用头文件<string.h>里的memset函数，它不仅可以全盘赋值，而且赋值自由。

#2 细节问题：因为Output a blank line between two test cases，所以最后一组数据后面没有空行