Problem:
Score : 300 points

Problem Statement

You are given a string S consisting of digits between 1 and 9, inclusive.
You can insert the letter + into some of the positions (possibly none) between two letters in this string.
Here, + must not occur consecutively after insertion.

All strings that can be obtained in this way can be evaluated as formulas.

Evaluate all possible formulas, and print the sum of the results.

Constraints

1 \leq |S| \leq 10

All letters in S are digits between 1 and 9, inclusive.

Input

The input is given from Standard Input in the following format:

S

Output

Print the sum of the evaluated value over all possible formulas.

Sample Input 1

125

Sample Output 1

176

There are 4 formulas that can be obtained: 125, 1+25, 12+5 and 1+2+5. When each formula is evaluated,

125

1+25=26

12+5=17

1+2+5=8

Thus, the sum is 125+26+17+8=176.

Sample Input 2

9999999999

Sample Output 2

12656242944

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<vector>
#include<utility>
#include<algorithm>
#include<map>
#include<cstdlib>
#include<cmath>
#include<numeric>
#include<iomanip>
#include<functional>
#include<cstdlib>
#include<queue>
#include<deque>
const double PI = acos(-1);
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
using ll =long long;
const int mod = 1000000007;

//map<string>a
class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% mod + mod) % mod) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint & a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint & operator-=(const mint & a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint & operator*=(const  mint & a) {
		(x *= a.x) %= mod;
		return *this;
	}
	mint operator+(const mint & a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint & a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint & a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}
	// for prime mod
	mint inv() const {
		return pow(mod - 2);
	}
	mint& operator/=(const mint & a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint & a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream & os, const mint & m) {
		os << m.x;
		return os;
	}
};
int ctoi(char c) {
	if (c >= '0' && c <= '9') {
		return c - '0';
	}
	return 0;
}
//組み合わせ
void recursive_comb(vector<int> indexes, int s, int rest, std::function<void(vector<int>)> f) {
	if (rest == 0) {
		f(indexes);
	}
	else {
		if (s < 0) return;
		recursive_comb(indexes, s - 1, rest, f);
		indexes[rest - 1] = s;
		recursive_comb(indexes, s - 1, rest - 1, f);
	}
}

// nCkの組み合わせに対して処理を実行する
void foreach_comb(int n, int k, std::function<void(vector<int>)> f) {
	vector<int>indexes(k);
	recursive_comb(indexes, n - 1, k, f);
}

ll gcd(ll a, ll b) {
	if (a % b == 0) {
		return b;
	}
	else {
		return gcd(b, a % b);
	}
}
ll lcm(ll a, ll b) {
	return a * b / gcd(a, b);
};
//fixed << setprecision(2)
ll number(char c, int i) {
	ll tmp = (ll)c;
	tmp = tmp - 48;
	tmp *= pow(10, i);
	return tmp;
}
int main() {
	string s;
	cin >> s;
	int n = s.size()-1;
	ll ans=0;
	int head;
	for (int bit = 0; bit < (1 << n); ++bit) {
		vector<int> S;
		for (int i = 0; i < n; ++i) {
			if (bit & (1 << i)) { // 列挙に i が含まれるか
				S.push_back(i);
			}
		}
		head = 0;
		for (int i = 0; i < (int)S.size(); ++i) {
			for (int j = head; j <= S[i];j++) {

				ans += number(s[j],S[i]-j);

			}
			head = S[i]+1;

		}
		for (int j = head; j <s.size(); j++) {
			ans += number(s[j], s.size()-1-j);
		}
		
	}
	cout << ans;
}