1.36进制转10进制（这题我没判断大小试着调试一下直接给A了…）
import java.util.Scanner;
public class Right1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String word = sc.next();
int length = word.length();
long result = 0;
if(length==0){
System.out.println(result);
}else if(length==1){
if((int)word.charAt(0)>=48&&(int)word.charAt(0)<=57){
result = (int)word.charAt(0)-48;
}else if((int)word.charAt(0)>=97&&(int)word.charAt(0)<=122){
result = (int)word.charAt(0)-87;
}else{
result = 0;
}
System.out.println(result);
}else {
for(int i =1;i<word.length();i++){
if((int)word.charAt(i)>=48&&(int)word.charAt(i)<=57){
result += Math.pow(36,word.length()-i-1)*((int)word.charAt(i)-48);
}else if((int)word.charAt(i)>=97&&(int)word.charAt(i)<=122){
result += Math.pow(36,word.length()-i-1)*((int)word.charAt(i)-87);
}else{
result = 0;
break;
}
}
if((int)word.charAt(0)==45){
result*=-1;
}
else if((int)word.charAt(0)>=48&&(int)word.charAt(0)<=57){
result += Math.pow(36,word.length()-1)*((int)word.charAt(0)-48);
}else if((int)word.charAt(0)>=97&&(int)word.charAt(0)<=122){
result += Math.pow(36,word.length()-1)*((int)word.charAt(0)-87);
}else{
result = 0;
}
System.out.println(result);
}
}
} 2.有序数组最多构建多少个二叉搜索树（动归，不同节点做根节点的情况下左子树组合数量*右子树组合数量） import java.util.Scanner;
public class Right2 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
if(n==0){
System.out.println(0);
}else{
int [] array = new int [n+1];
array[0] = 1;
for(int i =1;i<array.length;i++){
for(int j = 1;j<=i;j++){
array[i]+=array[j-1]*array[i-j];
}
}
System.out.println(array[n]); }
}
} 
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