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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

typedef long long ll;

int main() {
  int i, j;
  // s を文字列として読み込む
  char s[20];
  scanf("%s", s);

  int s_len = strlen(s);
 // printf("%d\n", s_len);

  int a[20];
  for (i = 0; i < s_len; i++) {
    a[i] = s[i]-'0';
  }

  for (i = 0; i < s_len; i++) {
   // printf("%d\n", a[i]);
  }

  ll ans = 0; // 答えは int に収まらないので long long int を使う
  // s_len 個の数字の間は s_len-1 個ある
  // それぞれの間に, + を入れるか入れないかを考える.
  for (i = 0; i < (1<<(s_len-1)); i++) {
    // i: 000..00 ~ 111..11
    // i 番目の bit が立っている => i番目の間に + を入れると考える
    ll cur = a[0], sum = 0;
    for (j = 0; j < s_len-1; j++) {
      // j 番目の bit が立っていたら
      // j 番目の間に入るのは +
      if ( (i>>j) & 1 ) {
        sum += cur;
        cur = a[j+1];
      } else {
        cur *= 10;
        cur += a[j+1];
      }
    }
    sum += cur;
    ans += sum;
  }

  printf("%lld\n", ans);

  return 0;
}