洛谷P1634、P1888、P3954、P5660题解(Java语言描述)
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洛谷P1634、P1888、P3954、P5660题解(Java语言描述)
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P1634題目要求
P1634題目鏈接
P1634題解
import java.util.Scanner;public class Main {public static void main(String[] args) {long sum = 1L;Scanner scan = new Scanner(System.in);int a = scan.nextInt(), b = scan.nextInt();scan.close();if (a > 0) {if (b > 0) {for (int i = 0; i < b; i++) {sum += sum*a;}}}System.out.println(sum);}}P1888題目要求
P1888題目鏈接
P1888題解
import java.util.Scanner;public class Main {public static void main(String[] args) {int maxNumber, minNumber, a;Scanner scan = new Scanner(System.in);int x = scan.nextInt(), y = scan.nextInt(), z = scan.nextInt();if (x >= y && x >= z) {maxNumber = x;if (y >= z)minNumber = z;elseminNumber = y;} else if (y >= x && y >= z) {maxNumber = y;if (x >= z)minNumber = z;elseminNumber = x;} else {maxNumber = z;if (x >= y)minNumber = y;elseminNumber = x;}int number1 = maxNumber;int number2 = minNumber;while(number2!=0){int r = number1 % number2;number1 = number2;number2 = r;}//在while過程中,a與b的值會發生改變,所以需要將a,b的值賦給oa,ob//oa與ob的最大公約數是aa = number1;minNumber /= a;maxNumber /= a;System.out.println(minNumber + "/" + maxNumber);scan.close();}}P3954題目要求
P3954題目鏈接
P3954題解
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scan = new Scanner(System.in);int x = scan.nextInt(), y = scan.nextInt(), z = scan.nextInt();int result = (int)(x*0.2 + y*0.3 + z*0.5);System.out.println(result);scan.close();}}P5660題目要求
P5660題目鏈接
P5660題解
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);String line = scanner.nextLine();scanner.close();char[] arr = line.toCharArray();int counter = 0;for (char c : arr) {if (c == 49) {counter++;}}System.out.println(counter);} }都是以前寫的,如果lj,還請見諒
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