1001. A+B Format (20)

最新推荐文章于 2021-02-19 15:05:15 发布

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最新推荐文章于 2021-02-19 15:05:15 发布

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本文链接：https://blog.youkuaiyun.com/heroacool/article/details/48937399

本文介绍如何将两个整数相加，并将其结果按照标准格式输出，即数字被分组为三位一节，用逗号隔开。通过示例输入和输出，详细解释了解题思路，包括字符串转换和特定的格式化步骤。

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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

解题思路：
1，转化为字符串处理。
2，转化为%3部分和整除3的部分

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


int main(){

    int a, b;
    cin >> a >> b;
    int sum = a + b;

    if (sum < 0){
        cout << "-";
        sum = -sum;
    }

    char buffer[10];
    sprintf(buffer, "%d", sum);

    string s(buffer);
    int k = s.size() / 3;
    int k1 = s.size() % 3;
    for (int i = 0; i < k1; i++){
        cout << s[i];
    }
    if (k1 != 0){
        for (int j = k1; j < s.size(); j++){
            int n = j - k1;
            if (n % 3 == 0){
                cout << ",";
            }
            cout << s[j];
        }
    }
    else{
        for (int j = 0; j < s.size(); j++){
            int n = j ;
            if (n % 3 == 0){
                if (j != 0)
                    cout << ",";
            }
            cout << s[j];
        }
    }




}

第二种思路：
1，转化为字符串，然后从末位开始计数，逢3，push一个’,’，然后反向字符串，判断若反向后的字符串首字符为’,’，则不输出即可。

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


int main(){

    int a, b;
    cin >> a >> b;
    int sum = a + b;

    if (sum < 0){
        cout << "-";
        sum = -sum;
    }

    char buffer[10];
    sprintf(buffer, "%d", sum);

    string s(buffer);
    //cout << s << endl;

    string temp_s;

    for (int i = s.size() - 1; i >= 0; i--){
        temp_s.push_back(s[i]);
        int counter = s.size() - i;
        if (counter % 3 == 0){
            temp_s.push_back(',');
        }
    }
    for (int i = 0,j = temp_s.size() - 1; i < j; i++, j--){
        char c = temp_s[i];
        temp_s[i] = temp_s[j];
        temp_s[j] = c;
    }

    if (temp_s[0] != ','){
        cout << temp_s[0];
    }
    for (int i = 1; i < temp_s.size(); i++)
        cout << temp_s[i];




}