PAT基础编程题目-6-9 统计个位数字

    技术2025-02-06  18

    PAT基础编程题目-6-9 统计个位数字

    题目详情

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

    解答

    C语言版

    #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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