思路：
直接java模拟递归写。
不过python貌似可以用语法直接写。
import java.util.*;
import java.math.*;
import java.io.*;
import java.lang.*;public class Main {public static BigInteger qpow(BigInteger n) {BigInteger res = BigInteger.ONE;BigInteger zero = BigInteger.ZERO;BigInteger one = BigInteger.ONE;BigInteger two = one.add(one);BigInteger x = two;while(n != zero) {if(n.mod(two).equals(one)) {res = res.multiply(x);}n = n.divide(two);x = x.multiply(x);}return res;}public static BigInteger cal(int l,int r,String s) {BigInteger zero = BigInteger.ZERO;BigInteger one = BigInteger.ONE;BigInteger two = one.add(one);if(s.charAt(l) == '(' && s.charAt(r) == ')') {l++;r--;}if(l == r) {if(s.charAt(l) == '0') return zero;else if(s.charAt(l) == '2') return two;}int pre = l;BigInteger res = BigInteger.ZERO;int cnt = 0;for(int i = l;i <= r;i++) {if(s.charAt(i) == '(') cnt--;else if(s.charAt(i) == ')') cnt++;else if(s.charAt(i) == '+' && cnt == 0) {BigInteger num = cal(pre,i - 1,s);
// System.out.println(num + " " + pre + " " + (i - 1));res = res.add(num);pre = i + 1;}}if(pre < r) pre++;BigInteger num = cal(pre,r,s);if(r - pre + 1 != 1) num = qpow(num);res = res.add(num);return res;}public static void main(String[] args) {Scanner cin = new Scanner(System.in);String s;s = cin.next();int len = s.length();BigInteger ans = cal(0,len - 1,s);System.out.println(ans);}
}//2(2+2(0))