Table of contents

• Problem
• Time complexity
• Solution

# Problem

https://atcoder.jp/contests/arc061/tasks/arc061_a

# Time complexity

$S$ is the length of the string.

$O(2^S)$

# Solution

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

#include <sstream>
string S;

void input() {
  cin >> S;
}

Int dfs(string s) {
  stringstream ss(s);
  Int x = 0;
  ss >> x;
  if (s.length() == 1) return x;
  loop(n,1,s.length()) {
    string left = s.substr(0, n);
    string right = s.substr(n);
    stringstream ss2(left); Int leftNum = 0; ss2 >> leftNum;
    Int rightNum = dfs(s.substr(n));
    // i.e. 1 + 23
    //      1 + 2 + 3
    // Here '1 +' is added twice. (twice is the number of cases)
    x += leftNum * pow(2, right.length()-1) + rightNum;
  }
  return x;
}

void solve() {
  cout << dfs(S) << endl;
}

int main(void) {
  input();
  solve();
  return 0;
}