A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3947 Accepted Submission(s): 1386

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

 

Author

Ignatius.L

这道题看似简单，实则是个大数相加问题，显然其位数已超过基本数据类型所能表示的最大大小。针对此题，基本思路如下：

1、将输入作为字符串保存，逐位进行单个字符的相加。

注意事项：

1、由于字符串的第一个字符表示数值的最高位，最后一个字符表示数值的最低位，因此在进行逐位相加前，应将两个加数先逆序，在相加。

2、相加计算时应该不要忽略进位以及两个加数位数不等的情况，还有就是相加的是字符的ASCII码值，因注意与字符‘0’做个数值转换哦。

3、输出的格式不对也不能AC的哦。

code：

#include <iostream>using namespace std;char a[1001] = { 0 };char b[1001] = { 0 };char a1[1001] = { 0 };char b1[1001] = { 0 };char sum[1002] = { 0 };int main(){    int n;    int cnt = 0;    int length_a1;    int length_b1;    cin >> n;    cnt = n;    while (n--)    {        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        memset(a1, 0, sizeof(a1));        memset(b1, 0, sizeof(b1));        memset(sum, 0, sizeof(sum));        cin >> a1;        cin >> b1;        // 逆序        length_a1 = strlen(a1);        int index = length_a1 - 1;        for (; index >= 0; index--)        {            a[length_a1 - 1 - index] = a1[index];        }        length_b1 = strlen(b1);        index = length_b1 - 1;        for (; index >= 0; index--)        {            b[length_b1 - 1 - index] = b1[index];        }        int i = 0;        int carry = 0;        while ((a[i] != 0) && (b[i] != 0)) // 两个都不为0        {            int tmp = (a[i] - '0') + (b[i] - '0') + carry;            if (tmp >= 10)            {                tmp = tmp - 10;                carry = 1;            }            else            {                carry = 0;            }            sum[i] = 0 + tmp + 48;            i++;        }        if(length_a1 > length_b1)        {            while (a[i] != 0)            {                int tmp = 0 + (a[i] - '0') + carry;                if (tmp >= 10)                {                    tmp = tmp - 10;                    carry = 1;                }                else                {                    carry = 0;                }                sum[i] = 0 + tmp + 48;                i++;            }        }        else        {            while (b[i] != 0)            {                int tmp = 0 + (b[i] - '0') + carry;                if (tmp >= 10)                {                    tmp = tmp - 10;                    carry = 1;                }                else                {                    carry = 0;                }                sum[i] = 0 + tmp + 48;                i++;            }        }        if (carry == 1)        {            sum[i] = 1 + 48;        }        cout << "Case " << cnt - n << ":" << endl;        cout << a1 << " + " << b1 << " = ";                // 逆序输出        int k = strlen(sum) - 1;        for (; k >= 0; k--)        {            cout << sum[k];        }        if (n == 0)        {            cout << endl;        }        else        {            cout << endl << endl;        }    }    return 0;}