乙级 - 1017 A除以B (20分)

本题要求计算 A / B，其中 A 是不超过 1000 位的正整数，B 是 1 位正整数。你需要输出商数 Q 和余数 R，使得 A = B × Q + R 成立。

输入格式

输入在一行中依次给出 A 和 B，中间以 1 空格分隔。

输出格式

在一行中依次输出 Q 和 R，中间以 1 空格分隔。

输入样例

123456789050987654321 7

输出样例

17636684150141093474 3

Java 代码

1. 运行超时（0 分）
   输入：Scanner in = new Scanner(System.in);
   输出：for i 循环

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        char[] numA = in.next().toCharArray();
        int numB = in.nextInt();

        int len = numA.length;
        int[] numQ = new int[len];
        int r = 0;
        for (int i = 0; i< len; i++) {
            int temp = numA[i] - '0';
            r = r * 10 + temp;
            numQ[i] = r / numB;
            r %= numB;
        }

        boolean tag = false;
        for (int i = 0; i < len; i++) {
        	if (len != 1) {
	        	if (numQ[i] != 0 && ) {
	                tag = true;
	            }
	            if (numQ[i] == 0 && !tag) {
	                continue;
	            }
        	}
            System.out.print(numQ[i]);
        }
        System.out.print(" " + r);
    }
}

2. 运行超时（0 分）
   输入：Scanner in = new Scanner(System.in);
   输出：System.out.print(result); 输出结果字符串

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        char[] numA = in.next().toCharArray();
        int numB = in.nextInt();

        StringBuilder numQ = new StringBuilder();
        int r = 0;
        for (char c : numA) {
            int temp = c - '0';
            r = r * 10 + temp;
            numQ.append(r / numB);
            r %= numB;
        }

        String result = numQ.toString() + " " + r;
        if (numQ.charAt(0) == '0' && numQ.length() != 1) {
            System.out.print(result.substring(1));
        } else {
            System.out.print(result);
        }
    }
}

3. 部分正确（18 分，测试点 4：运行超时）
   输入：BufferedReader、InputStreamReader
   输出：System.out.print(result); 输出结果字符串

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        String[] datas = bufferedReader.readLine().split(" ");
        char[] numA = datas[0].toCharArray();
        int numB = Integer.parseInt(datas[1]);

        StringBuilder numQ = new StringBuilder();
        int r = 0;
        for (char c : numA) {
            int temp = c - '0';
            r = r * 10 + temp;
            numQ.append(r / numB);
            r %= numB;
        }

        String result = numQ.toString() + " " + r;
        if (numQ.charAt(0) == '0' && numQ.length() != 1) {
            System.out.print(result.substring(1));
        } else {
            System.out.print(result);
        }
    }
}

4. AC（20 分）
   与 3 唯一不同的是，处理数据时，采用 for i 循环，避免使用了 foreach 循环。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        String[] datas = bufferedReader.readLine().split(" ");
        char[] numA = datas[0].toCharArray();
        int numB = Integer.parseInt(datas[1]);
        
        StringBuilder numQ = new StringBuilder();
        int r = 0;
        for (int i = 0, len = numA.length; i < len; i++) {
            int temp = numA[i] - '0';
            r = r * 10 + temp;
            numQ.append(r / numB);
            r %= numB;
        }

        String result = numQ.toString() + " " + r;
        if (numQ.charAt(0) == '0' && numQ.length() != 1) {
            System.out.print(result.substring(1));
        } else {
            System.out.print(result);
        }
    }
}

C/C++ 代码

#include <iostream>
using namespace std;

int main() {
    string numA;
    int numB, r = 0;
    cin >> numA >> numB;
    
    int numQ[1005];
    int cnt = 0;
    for (int i = 0, len = numA.length(); i < len; i++) {
        int temp = numA[i] - '0';
        r = r * 10 + temp;
        numQ[cnt++] = (r / numB);
        r %= numB;
    }

	bool tag = false;
	for (int i = 0; i < cnt; i++) {
		if (numA.length() != 1) {
			if (numQ[i] != 0) {
				tag = true;
			} else if (numQ[i] == 0 && !tag ) {
				continue;
			}	
		}
		cout << numQ[i];	
	}

	cout << " " << r;
    return 0;
}