PAT基礎編程題目-6-9 統計個位數字

PAT基礎編程題目-6-9 統計個位數字

題目詳情

題目地址：https://pintia.cn/problem-sets/14/problems/741

解答

C語言版

#include <stdio.h>

int Count_Digit(const int N, const int D);

int main()
{
	int N, D;

	scanf("%d %d", &N, &D);
	printf("%d\n", Count_Digit(N, D));
	return 0;
}

int Count_Digit(const int N, const int D) {
	int n = N;
	int number[10];
	int size = 0;
	int count = 0;
	for (int i = 0; i < 10; i++)
	{
		number[i] = n % 10;
		printf("number[%d]=%d\n", i, number[i]);
		n = n / 10;
		size++;
		if (n == 0) {
			break;
		}
	}
	for (int j = 0; j < size; j++) {
		if ((D + number[j] == 0) || (D - number[j]== 0)) {
			count++;
		}
	}
	return count;
}

C++版

#include<iostream>
using namespace std;
int Count_Digit(const int N, const int D);

int main()
{
	int N, D;

	cin >> N >> D;
	cout<<Count_Digit(N, D);
	return 0;
}
int Count_Digit(const int N, const int D) {
	int n = N;
	int number[10];
	int size = 0;
	int count = 0;
	for (int i = 0; i < 10; i++)
	{
		number[i] = n % 10;
		n = n / 10;
		size++;
		if (n == 0) {
			break;
		}
	}
	for (int j = 0; j < size; j++) {
		if ((D + number[j] == 0) || (D - number[j] == 0)) {
			count++;
		}
	}
	return count;
}

Java版

public class Main{

	private static int Count_Digit(final int N, final int D) {
		int n = N;
		int [] number = new int[10];
		int size = 0;
		int count = 0;
		for (int i = 0; i < 10; i++)
		{
			number[i] = n % 10;
			n = n / 10;
			size++;
			if (n == 0) {
				break;
			}
		}
		for (int j = 0; j < size; j++) {
			if ((D + number[j] == 0) || (D - number[j] == 0)) {
				count++;
			}
		}
		return count;
	}
	
	public static void main(String[] args) {
		int N=0,D=0;
		Scanner scanner = new Scanner(System.in);
		if(scanner.hasNext()) {
			N = scanner.nextInt();
			D = scanner.nextInt();
		}
		scanner.close();
		System.out.println(Count_Digit(N, D));

	}

}

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