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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <numeric>
//#include <stdio.h>
//#include <stack>
//#include <queue>
//#include <cstdio>
#include <cmath>
#include <iterator>
#include <map>
#include <fstream>
//#include <list>
#include <iomanip>
//#include <stdlib.h>
//#include <cctype>

using namespace std;

#define IOS ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define prec std::cout << std::fixed << std::setprecision(15);
#define endl "\n"
#define MOD 1000000007
#define Int int64_t
#define PI 3.14159265358979
#define ssort(z) sort(z.begin(), z.end())
#define rsort(z) sort(z.rbegin(), z.rend())
#define eerase(z) z.erase(unique(z.begin(), z.end()), z.end())
#define ccnt(z, w) count(z.begin(), z.end(), w)
#define rep(i,a,n) for(Int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(Int (i)=(a); (i)<=(n); (i)++)
const int MAX_N = 1000000;
const Int MAX_N_Int = 1000000000000;

template <typename T>
void printV(const std::vector<T>& v, const char * const separator = " ")
{
    if(!v.empty())
    {
        std::copy(v.begin(),
                  --v.end(),
                  std::ostream_iterator<T>(std::cout, separator));
        std::cout << v.back() << "\n";
    }
}

bool isPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}
Int gcd(Int a, Int b) {
    return b != 0 ? gcd(b, a % b) : a;
}

Int lcm(Int a, Int b) {
    return a / gcd(a, b) * b;
    //a*bは64bit integer overflow
}

int Max(int a, int b, int c) {
    int temp = max(a, b);
    return max(temp, c);
}

int Min(int a, int b, int c) {
    int temp = min(a, b);
    return min(temp, c);
}

bool integer(double num) {
    return floor(num) == num;
}

Int fact(int num) {
    if (num == 0)
        return 1;
    else
        return num * fact(num - 1);
}

Int yakusu(int n) {
    int cnt = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            cnt++;
            if (i * i != n) cnt++;
        }
    }
    return cnt;
}


Int fact_mod(Int n, Int mod) {
    Int f = 1; repq(i,2,n) f = f * (i % MOD) % MOD;
    return f;
}

// 繰り返し二乗法 (modの世界での累乗)
Int mod_pow(Int x, Int n, Int mod) {
    Int res = 1;
    while(n > 0) {
        if(n & 1) res = (res * x) % mod;
        x = (x * x) % mod;
        n >>= 1;
    }
    return res;
}

// 組み合わせ nCr を求める
Int combination_mod(Int n, Int r, Int mod) {
    if(r > n-r) r = n-r;
    if(r == 0) return 1;
    Int a = 1;
    rep(i, 0, r) a = a * ((n-i) % mod) % mod;
    Int b = mod_pow(fact_mod(r, mod), mod-2, mod);
    return (a % mod) * (b % mod) % mod;
}

bool rev(string s){
    string t=s;
    reverse(s.begin(),s.end());
    return s==t;
}
string s;
Int ans=0,tmp=0,n,sum=0;
vector<bool>p;
int dfs(int now,vector<bool>&pl){
    printV(pl);
    if(now==n-1){
        bool prevPlus=false;
        ans=0;
        for (int i = 0; i < n; ++i) {
            if(i==n-1){
                ans+=(tmp+s[i]-'0');
                break;
            }
            if(pl[i]){
                if(prevPlus){
                    ans+=s[i]-'0';
                }else{
                    ans+=(tmp+s[i]-'0');
                    tmp=0;
                }

            }
            else{
                tmp*=10;
                tmp+=10*(s[i]-'0');
            }
            prevPlus=pl[i];
        }
        sum+=ans;
    }else{
        pl[now]=true;
        dfs(now+1,pl);
        pl[now]=false;
        dfs(now+1,pl);

    }
    return 0;
}
int dfs2(int now,vector<bool>&pl){
    //printV(pl);
    if(now==n-1){
        bool prevPlus=false;
        ans=0;
        tmp=0;
        for (int i = 0; i < n; ++i) {
            if(i==n-1){
                ans+=(tmp+s[i]-'0');
                break;
            }
            if(pl[i]){
                if(prevPlus){
                    ans+=s[i]-'0';
                }else{
                    ans+=(tmp+s[i]-'0');
                    tmp=0;
                }

            }
            else{
                tmp*=10;
                tmp+=10*(s[i]-'0');
            }
            prevPlus=pl[i];
        }
        //cout<<ans<<endl;
        sum+=ans;
    }
    return 0;
}
int main() {
    //COMinit();
    IOS;
    prec;
    cin>>s;
    n=s.length();
    p.resize(n-1);
    //dfs(0,p);
    //cout<<sum;

    p.resize(n-1);
    for (int bit = 0; bit < (1<<n-1); ++bit) {
        for (int i = 0; i < n; ++i) {
            if (bit & (1<<i)) { // 列挙に i が含まれるか
                p[i]=true;
            }else{
                p[i]=false;
            }
        }
        dfs2(n-1,p);

    }
    cout<<sum;









    cout<<endl;
    return 0;
}

/*
テレビも無ェ ラジオも無ェ
自動車もそれほど走って無ェ
ピアノも無ェ バーも無ェ
巡査 毎日ぐーるぐる
朝起ぎで 牛連れで
二時間ちょっとの散歩道
電話も無ェ 瓦斯も無ェ
バスは一日一度来る
俺らこんな村いやだ 俺らこんな村いやだ
東京へ出るだ 東京へ出だなら
銭コァ貯めで 東京でベコ(牛)飼うだ
*/