1002 A + B Problem II - 杭电 Oj | 方家小白 = 和你一起遇见更好的自己

# 题目信息

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 565951 Accepted Submission(s): 107979

# 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

1
2
3

2
1 2
112233445566778899 998877665544332211

# Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

# 题解

# C 语言解法

#include <stdio.h>

int main() {

    int n, i = 0;
    scanf("%d", &n);
    char num1[1001];
    char num2[1001];
    while (i < n) {
        char res[1001];
        int res_end_pos = 0;
        int temp = 0;
        int num1_end_pos = 0;
        int num2_end_pos = 0;

        scanf("%s %s", num1, num2);

        // 将下标计算到数值的个位。
        while ('\0' != num1[num1_end_pos]) {
            num1_end_pos++;
        }

        while ('\0' != num2[num2_end_pos]) {
            num2_end_pos++;
        }

        // 按照下标从大到小的顺序，累加，并计算是否需要进位。
        res_end_pos = (num1_end_pos > num2_end_pos ? num1_end_pos : num2_end_pos) + 1;
        res[res_end_pos] = '\0';
        while (num1_end_pos > 0 && num2_end_pos > 0) {
            num2_end_pos--;
            num1_end_pos--;
            res_end_pos--;

            int number1 = num1[num1_end_pos] - '0';
            int number2 = num2[num2_end_pos] - '0';
            temp = number1 + number2 + (temp / 10);
            res[res_end_pos] = (temp % 10) + '0';
        }

        // 计算进位以及某个数值位数不一致的情况
        while (res_end_pos > 0) {
            res_end_pos--;
            int temp_res = 0;
            if (num1_end_pos == 0 && num2_end_pos > 0) {
                num2_end_pos--;
                temp_res = (num2[num2_end_pos] - '0') + temp / 10;
            } else if (num1_end_pos > 0 && num2_end_pos == 0) {
                num1_end_pos--;
                temp_res = (num1[num1_end_pos] - '0') + temp / 10;
            } else {
                temp_res = temp / 10;
            }
            temp = temp_res;
            res[res_end_pos] = temp_res % 10 + '0';
        }
        i++;

        // 处理没有进位导致的输出问题(多输出一个0)
        if(res[0] == '0'){
            int j=0;
            while(res[j]!='\0'){
                res[j] = res[j+1];
                j++;
            }
        }

        printf("Case %d:\n", i);
        printf("%s + %s = %s\n", num1, num2, res);
        if(i < n){
            printf("\n");
        }
    }

    return 0;
}

# Be Careful

注意输出结果：每个 case 之前会隔一个空行。最后一个 case 后面没有空行。注意读题。

# 长点心吧

理清需求啊

# 链接

杭电Oj - 1002

http://acm.hdu.edu.cn/

【C】杭电Oj-1002题解

https://gitee.com/fangjiaxiaobai/leetCode.git

# 最后

希望与你一起遇见更好的自己

期望与你一起遇见更好的自己

HD BQ AC