A + B Problem II（HDOJ_1002）

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
	

	

		
	

 

     题目解析:
        与HDOJ_1000不同，这里包含大整数加法。因此，不能按那个方法进行处理，这里采用了字符串进行大整数的加法。
    代码如下：

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

#define  MAXNUM 1004

//交换
void Swap(char *a, char *b)
{
    char tmp = *a;
    *a = *b;
    *b = tmp;
}
//反转
void reserve(char *a, int len)
{
    int i, j;
    if(a == NULL || len == 0)
    {
        return;
    }

    i = 0;
    j = len - 1;
    while(i < j)
    {
        Swap(&a[i], &a[j]);
        i++;
        j--;
    }
}

int main(int argc, char* argv[])
{
    int t = 0;
    char a[MAXNUM];
    char b[MAXNUM];
    int c[MAXNUM];
    int i = 0, j = 1;
    int count = 1;
    int len_a, len_b, len_c;

    scanf("%d", &t);
    if(t <= 0 || t >= 21)
    {
        return 0;
    }
    for(j = 1; j <= t; j++)
    {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        memset(c, 0, sizeof(c));

        scanf("%s %s", a, b);

        printf("Case %d:\n", count++);
        printf("%s + %s = ", a, b);

        len_a = strlen(a);
        len_b = strlen(b);
        len_c = (len_a > len_b) ? len_a : len_b;

        //反转a和b
        reserve(a, len_a);
        reserve(b, len_b);

        //转为数字
        for(i = 0; i < len_a; i++)
        {
            a[i] -= '0';
        }
        for(i = 0; i < len_b; i++)
        {
            b[i] -= '0';
        }

        //计算
        for(i = 0; i < len_c; i++)
        {
            c[i] = c[i] + a[i] + b[i];
            c[i + 1] += c[i] / 10;
            c[i] %= 10;
        }

        //有进位
        if(c[len_c] > 0)
        {
            len_c++;
        }

        //输出
        for(i = len_c - 1; i >= 0; i--)
        {
            printf("%d",c[i]);
        }
        printf("\n");
        if(j != t)
        {
            printf("\n");
        }
    }
    return 0;
}