PAT: 1001. A+B Format (20)

最新推荐文章于 2025-03-05 20:47:12 发布

原创最新推荐文章于 2025-03-05 20:47:12 发布 · 314 阅读

0 ·

CC 4.0 BY-SA版权

pat 专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一种方法，用于解决两个整数相加后如何按照标准格式输出的问题，即在千位数上使用逗号分隔。通过将结果转换为字符串并逆向遍历来实现在适当位置插入逗号的功能。

问题描述：
Calculate a + b and output the sum in standard format – that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input
-1000000 9
Sample Output
-999,991

思路：先进行普通的A+B，将得到的结果转换成字符串处理，从后往前遍历加入一个空串，每隔插入一个逗号，注意最高位的数字前不要加逗号。
具体代码如下：

#include <iostream>
#include <sstream>
#include <string>
using namespace std;

int main()
{
    int a, b, s;
    cin >> a >> b;
    int c = a + b;
    stringstream ss;
    ss << c;
    string t;
    ss >> t;
    if(t[0] == '-')
        s = 1;
    else
        s = 0;
    string str = "";
    int k = 0;
    for(int i = t.length() - 1; i >= s; i --) {
        str = t[i] + str;
        k ++;
        if(k == 3 && i != s) {
            str = ',' + str;
            k = 0;
        }
    }
    if(s == 1)
        str = '-' + str;
    cout << str << endl;
    return 0;
}