我正在开发一个游戏，很像数学骰子问题，尽管有点不同。用户掷 20 面骰子，然后再掷 5 个骰子。为了让事情变得更简单，用户无法重新排序骰子，因此如果他们掷 1 2 3 4 5，他们就无法执行 1 + 3 + 2 + 5 + 4 等操作。问题是，使用加法、减法和乘法，他们能否从 20 面骰子中达到目标数字？
现在，我知道如何做到这一点，只需生成 5 个数字的所有可能的加法、减法和乘法的排列，但让我着迷的是解决方案的实现。经过几次尝试后我遇到了障碍，因此感谢您的帮助。
编辑:这是我当前的实现，没有乘法，并且工作得不太正确。
import java.util.ArrayList;
import java.util.Scanner;
public class targetDice {
public static void main(String[] args) {
ArrayList<Integer> rolls = new ArrayList<Integer>(); // Array to hold the rolls
ArrayList<Integer> d20 = new ArrayList<Integer>(); // Array to hold all the d20 rolls
Scanner sc = new Scanner(System.in);
int answer = 0;
String record = "";
while (sc.hasNextInt()) {
d20.add(sc.nextInt()); // Adds the d20 rolls
rolls.add(sc.nextInt()); // Adds the first roll
rolls.add(sc.nextInt()); // Adds the second roll
rolls.add(sc.nextInt()); // Adds the third roll
rolls.add(sc.nextInt()); // Adds the fourth roll
rolls.add(sc.nextInt()); // Adds the fifth roll
} // End while loop
for (int i = 0; i < d20.size(); i++) { // Number of times we need to compute: number of d20 rolls
answer = rolls.get(0);
for (int j = 0; j < rolls.subList(0, 5).size(); j++) { // Go through each roll given
if (d20.get(i) > answer || d20.get(i).equals(answer)) { // If the d20 roll is higher than the first roll or if it's equal
answer += rolls.get(j);// then take the running total and add it
record += " + ";
} else if (d20.get(i) < answer) {
answer -= rolls.get(j);
record += " - ";
}
}
System.out.println(answer);
//TODO: This if else block is our final product. It should be fine.
if (answer == d20.get(i)) // If the combo is equal the d20 roll
System.out.println("Solution"); // Print solution
else
System.out.println("No Solution"); // Otherwise print no solution
rolls.subList(0, 5).clear(); // Clears out the first 5 elements to make coding easier
answer = 0; // Reset the answer var
System.out.println(record);
} // End For loop
} // End main
} // End class
它的设置是为了让用户可以多次进行掷骰，如果他们尝试这个游戏 3 次，他们可以完成所有这三个操作，然后一次获得所有三个答案。
如果你想以不同的方式查看它，这里是粘贴箱:http://pastebin.com/PRB0NKpN
编辑2:这是我的最终解决方案。有点武力。
import java.util.ArrayList;
import java.util.Scanner;
public class testClass {
public static void main(String[] args) {
ArrayList<Integer> d20 = new ArrayList<Integer>();
ArrayList<Integer> rolls = new ArrayList<Integer>();
Scanner sc = new Scanner(System.in);
while (sc.hasNextInt()) {
d20.add(sc.nextInt());
rolls.add(sc.nextInt());
rolls.add(sc.nextInt());
rolls.add(sc.nextInt());
rolls.add(sc.nextInt());
rolls.add(sc.nextInt());
}
int num1 = 0, num2 = 0, num3 = 0, num4 = 0;
for (int x = 0; x < d20.size(); x++) {
int wright = 0, rong = 0;
for (int i = 1; i < 4; i++) {
for (int j = 1; j < 4; j++) {
for (int k = 1; k < 4; k++) {
for (int m = 1; m < 4; m++) {
if (i == 1) {
num1 = rolls.get(0) + rolls.get(1);
} else if (i == 2) {
num1 = rolls.get(0) - rolls.get(1);
} else if (i == 3) {
num1 = rolls.get(0) * rolls.get(1);
}
if (j == 1) {
num2 = num1 + rolls.get(2);
} else if (j == 2) {
num2 = num1 - rolls.get(2);
} else if (j == 3) {
num2 = num1 - rolls.get(2);
}
if (k == 1) {
num3 = num2 + rolls.get(3);
} else if (k == 2) {
num3 = num2 - rolls.get(3);
} else if (k == 3) {
num3 = num2 * rolls.get(3);
}
if (m == 1) {
num4 = num3 + rolls.get(4);
} else if (m == 2) {
num4 = num3 - rolls.get(4);
} else if (m == 3) {
num4 = num3 * rolls.get(4);
}
if (d20.get(x) == num4) {
wright = 1;
}
}
}
}
}
if (wright == 1)
System.out.println("Case " + (x+1) + ": Solution");
else
System.out.println("Case " + (x+1) + ": No Solution");
rolls.subList(0, 5).clear();
}
}
}
最佳答案
我看到你自己找到了答案，但我也尝试解决你的问题，无论如何我决定在这里发布另一个解决方案:
import java.util.ArrayList;
import java.util.Scanner;
public class Test {
public static void main(String[] args){
Test test = new Test();
test.combineOperators();
Scanner scanner = new Scanner(System.in);
int result = scanner.nextInt(); //get input
int[] numbers = new int[5];
for(int i = 0; i <5; i++){
numbers[i] = scanner.nextInt();
}
ArrayList<Integer> results = test.operationsOnArrays(numbers, test.combineOperators()); //check for results
if(results.contains(result)){
System.out.println(result + " is a possible solution");
}else{
System.out.println(result + " is not a possible solution");
}
}
public ArrayList<String[]> combineOperators(){ //create all possible combinations of operators
String[] signs = {"+","-","*"};
ArrayList<String[]> combinations = new ArrayList<String[]>();
for(String a : signs){
for (String b : signs){
for(String c : signs){
for(String d: signs){
String[]temp = {a,b,c,d};
combinations.add(temp);
}
}
}
}
return combinations;
}
public ArrayList operationsOnArrays(int[] num, ArrayList<String[]> combinations){ //do the math with every combination on given ints
ArrayList<Integer> list = new ArrayList<Integer>();
for(String[] operators : combinations){ //for every operators combination
int result = num[0];
for(int i = 0; i<=3 ; i++){
result = doTheMath(operators[i],result,num[i+1]); // take two ints and operator
}
list.add(result);
}
return list;
}
public int doTheMath(String operator, int prev, int next){ // it take two ints from input array, and do operation
if(operator.equals("+")){ // determined by a taken operator
return prev + next;
}else if(operator.equals("-")){
return prev - next;
}else if(operator.equals("*")){
return prev *next;
}
return 0;
}
}
我认为这样，扩展起来很简单，可以扩展更多的数字或运算符，甚至可以实现输入数字的重新排序。
关于java - 使用加法、减法或乘法可以使 5 个数字等于目标数字吗，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/30008414/