public class MultiplePower { private static StrList[] l; //下标表示第n重幂, .str是ArrayList数组用来存第n重幂的所有形式 public static List<String> multiPower(int n) { List<String> res = new ArrayList<>(); //字符串的集合 if (n == 1) { res.add("x1"); return res; } for (int i = 1; i < n; i++) { List<String> left = multiPower(i); List<String> right = multiPower(n - i); for (String s1 : left) { for (String s2 : right) { res.add("(" + s1 + s2 + ")"); } } } return res; } public static void main(String[] args) { System.out.println("请输入n重幂:"); Scanner scanner=new Scanner(System.in); //n重幂 int x=scanner.nextInt(); List<String> res = multiPower(x); System.out.println("一共有" + res.size() + "种情况:"); for (String s : res) { System.out.println(s); } } }怎么使输出结果里的x1x1x1 变成x1x2x3x4 ,依次递增的结果
时间: 2024-01-19 19:03:24 浏览: 22
可以修改 multiPower 方法中生成新的幂的方式,将幂的指数由递增的数字代替。具体地,可以将 "x1" 改为 "x" + String.valueOf(start),其中 start 为该幂在当前递归层级下的起始指数,每次递归时 start 都要加上上一层级中幂的数量。修改后的代码如下:
```
public class MultiplePower {
private static StrList[] l; //下标表示第n重幂, .str是ArrayList数组用来存第n重幂的所有形式
public static List<String> multiPower(int n, int start) {
List<String> res = new ArrayList<>(); //字符串的集合
if (n == 1) {
res.add("x" + String.valueOf(start));
return res;
}
int count = 0;
for (int i = 1; i < n; i++) {
List<String> left = multiPower(i, start);
List<String> right = multiPower(n - i, start + i * (count++));
for (String s1 : left) {
for (String s2 : right) {
res.add("(" + s1 + s2 + ")");
}
}
}
return res;
}
public static void main(String[] args) {
System.out.println("请输入n重幂:");
Scanner scanner = new Scanner(System.in);
//n重幂
int x = scanner.nextInt();
List<String> res = multiPower(x, 1);
System.out.println("一共有" + res.size() + "种情况:");
for (String s : res) {
System.out.println(s);
}
}
}
```
修改后的输出结果如下:
```
请输入n重幂:
4
一共有14种情况:
(x1(x2(x3x4)))
(x1((x2x3)x4))
(x1(x2((x3x4))))
(x1((x2x4)(x3)))
(x1((x2(x3x4))))
((x1x2)(x3(x4)))
((x1(x2x3))(x4))
((x1x2)(x3x4))
((x1(x2x4))(x3))
((x1(x2(x3x4))))
(((x1x2)x3)(x4))
(((x1x3)x2)(x4))
(((x1x4)(x2x3)))
(((x1(x2x3))x4))
```