我正在嘗試制作一個(gè)程序，告訴您進(jìn)行零錢所需的最少硬幣數(shù)量，并且它必須是模塊化的。目前它要求輸入三次并打印所有三個(gè)輸入的總和所需的硬幣數(shù)量。我相信這是因?yàn)?maxDimes 依賴于依賴于輸入的 maxQuarters。有什么辦法可以讓這些方法只運(yùn)行一次？我是編程和這個(gè)網(wǎng)站的新手，所以如果這個(gè)問(wèn)題不清楚或已經(jīng)被問(wèn)到，我深表歉意。 public static int input(){ System.out.println("Please enter amount of change under 100 cents: "); Scanner input = new Scanner(System.in); int change = input.nextInt(); if (change%5<2) { change = change+(change%5); } else if(change%5>2) { change = change-(change%5); } return change;}public static int[] maxQuarters(){ int change = input(); int quarters = change/25; int leftover = change-quarters; return new int[] {quarters, leftover};}public static int[] maxDimes(){ int first[] = maxQuarters(); int change = first[1]; int dimes = change/10; int leftover = change-dimes; return new int[] {dimes, leftover}; }public static int maxNickels(){ int second[] = maxDimes(); int change = second[1]; int nickels = change/5; return nickels;}public static void makeChange(){ int first[] = maxQuarters(); int quarters = first[0]; int second[] = maxDimes(); int dimes = second[0]; int nickels = maxNickels(); System.out.println("The minimum number of coins you could use is: "+quarters+"quarters, "+dimes+" dimes and "+nickels+" nickels");}public static void main(String[] args) { makeChange();}
1 回答
HUX布斯
TA貢獻(xiàn)1876條經(jīng)驗(yàn) 獲得超6個(gè)贊
您被要求輸入三遍，因?yàn)槟鷌nput()在maxQuarters(). 在maxDimes()和maxNickels()你叫maxQuarters()等也input()叫。
解決此問(wèn)題的一種可能方法是調(diào)用input()您的main-method 并將輸入作為參數(shù)傳遞給您的其他方法，如下所示：
public static int input() {
System.out.println("Please enter amount of change under 100 cents: ");
Scanner input = new Scanner(System.in);
int change = input.nextInt();
if (change % 5 < 2) {
change = change + (change % 5);
} else if (change % 5 > 2) {
change = change - (change % 5);
}
return change;
}
public static int[] maxQuarters(int input) {
int quarters = input / 25;
int leftover = input - 25*quarters;
return new int[]{quarters, leftover};
}
public static int[] maxDimes(int input) {
int first[] = maxQuarters(input);
int change = first[1];
int dimes = change / 10;
int leftover = change - 10*dimes;
return new int[]{dimes, leftover};
}
public static int maxNickels(int input) {
int second[] = maxDimes(input);
int change = second[1];
int nickels = change / 5;
return nickels;
}
public static void makeChange(int input) {
int first[] = maxQuarters(input);
int quarters = first[0];
int second[] = maxDimes(input);
int dimes = second[0];
int nickels = maxNickels(input);
System.out.println("The minimum number of coins you could use is: " + quarters + " quarters, " + dimes + " dimes and " + nickels + " nickels");
}
public static void main(String[] args) {
int input = input();
makeChange(input);
}
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