文章目錄

• • Java版
  • C++版

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 Specification:
Each input file contains one test case. Each case contains a pair of integers a and b where ?10^?6
?? ≤a,b≤10⁶??. The numbers are separated by a space.

Output Specification:
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

一開始做題的時候搞錯了一個點，導致測試點一半正確，一半錯誤：就是從前往后每3個數出一個逗號。就像給的測試用例一樣，但是這種方法只對字符串長度是3的整數倍的sum有效，要改成從后往前每3個數輸出一個逗號
用一個特殊情況舉例：
1000000+1000000=2000000

Java版

//20分 import java.io.*; import java.math.BigInteger;public class Main {public static void main(String[] args) throws IOException {//定義輸入BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));//輸入兩個整數String[] str = bf.readLine().split("\\s+");//將輸入的字符串轉化為BigInteger類型BigInteger a = new BigInteger(str[0]);BigInteger b = new BigInteger(str[1]);//計算兩個整數加和并轉化為String類型String sum = a.add(b) + "";//如果結果是負數，輸出“-”，之后將字符串中的“-”去掉if (sum.charAt(0) == '-') {System.out.print("-");sum = sum.substring(1);}//遍歷新字符串for (int i = 0; i < sum.length(); i++) {System.out.print(sum.charAt(i));//(i + 1) % 3 == sum.length() % 3是用來判斷逗號要加的位置//(i != sum.length() - 1))用來保證輸出的最后一位不輸出逗號if ((i + 1) % 3 == sum.length() % 3 && (i != sum.length() - 1)) {System.out.print(",");}continue;}} }

C++版

//參考柳神代碼 //沒想到思路和我的一樣 🤭 #include <iostream> #include<cstdio>using namespace std;int main() {int a, b;scanf("%d %d", &a, &b);string s = to_string(a + b);for (int i = 0; i < s.length(); i++) {cout << s[i];if (s[i] == '-')continue;if ((i + 1) % 3 == s.length() % 3 && i != s.length() - 1)cout << ",";}return 0; }

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