PAT基础编程题目-7-19 支票面额

PAT基础编程题目-7-19 支票面额

题目详情

题目详情：https://pintia.cn/problem-sets/14/problems/799

解答

C语言版

#include
int main() {
	int y, f, n;
	scanf("%d", &n);
	for (y = 0; y < 50; y++) {
		for (f = 0; f < 100; f++) {
		    // 仔细读懂题目，可以推出下面的公式
			if (98 * f - 199 * y - n == 0) {
				printf("%d.%d", y, f);
				return 0;
			}
		}
	}
	printf("No Solution");
	return 0;
}

C++版

#include
using namespace std;
int main() {
	int y, f, n;
	cin >> n;
	for (y = 0; y < 50; y++) {
		for (f = 0; f < 100; f++) {
			if (98 * f - 199 * y - n == 0) {
				cout << y << "." << f;
				return 0;
			}
		}
	}
	cout << "No Solution";
	return 0;
}

Java版

import java.util.Scanner;
public class Main{

	public static void main(String[] args) {
		int y, f, n = 0;
		Scanner scanner = new Scanner(System.in);
		if (scanner.hasNext()) {
			n = scanner.nextInt();
		}
		scanner.close();
		for (y = 0; y < 50; y++) {
			for (f = 0; f < 100; f++) {
				if (98 * f - 199 * y - n == 0) {
					System.out.println(y+"."+f);
					return;
				}
			}
		}
		System.out.println("No Solution");
	}

}

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