Problem:
Score : 300 points

Problem Statement

You are given an integer N. Among the integers between 1 and N (inclusive), how many Shichi-Go-San numbers (literally "Seven-Five-Three numbers") are there?

Here, a Shichi-Go-San number is a positive integer that satisfies the following condition:

When the number is written in base ten, each of the digits 7, 5 and 3 appears at least once, and the other digits never appear.

Constraints

1 \leq N < 10^9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go-San numbers between 1 and N (inclusive).

Sample Input 1

575

Sample Output 1

4

There are four Shichi-Go-San numbers not greater than 575: 357, 375, 537 and 573.

Sample Input 2

3600

Sample Output 2

13

There are 13 Shichi-Go-San numbers not greater than 3600: the above four numbers, 735, 753, 3357, 3375, 3537, 3557, 3573, 3575 and 3577.

Sample Input 3

999999999

Sample Output 3

26484

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int pow_(int a, int n){
  if(n == 0) return 1;
  else{
    return a * pow_(a, n - 1);
  }
}

int g(int n){
  if(n > 3) return 3 * g(n - 1) + pow_(2, n - 1) * 3 - 6;
  if(n == 3) return 6;
}

int
main(int argc, char *argv[])
{
  int n;
  scanf("%d", &n);

  int ans = 0, s, f, t, z, temp, x = 0;

  if(n <= 999){
    for(int l = 1; l <= n; l++){
      temp = l;
      s = 0; f = 0; t = 0; z = 0;
      while(temp > 0){
	if(temp % 10 == 7) s++;
        else if(temp % 10 == 5) f++;
        else if(temp % 10 == 3) t++;
        else { z++; break;}
        temp /= 10;
      }
      if(s*t*f > 0 && z == 0) ans++;
    }
    printf("%d\n", ans);
  }
  else{
    int d = 0;
    temp = n; int r;
    while(temp > 0){
      temp /= 10;
      r = temp % 10;
      d++;
    }
    for(int i = 3; i < d; i++){
      ans += g(i);
    }
    if(r == 3){
      for(int i = 3 * pow_(10, d - 1); i <= n; i++){
	s = 0; t = 0; f = 0; z = 0;
	temp = i;
	while(temp > 0){
	  if(temp % 10 == 3) t++;
	  else if(temp % 10 == 5) f++;
	  else if(temp % 10 == 7) s++;
	  else {z++; break;}
	}
	if(s*t*f > 0 && z == 0)ans++;
      }
    }
    else if(r == 4) ans += pow_(2, d - 1) + g(d - 1);
    else if(f == 5){
      for(int i = 5 * pow_(10, d - 1); i <= n; i++){
	s = 0; t = 0; f = 0; z = 0;
	temp = i;
	while(temp > 0){
	  if(temp % 10 == 3) t++;
	  else if(temp % 10 == 5) f++;
	  else if(temp % 10 == 7) s++;
	  else {z++; break;}
	}
	if(s*t*f > 0 && z == 0)ans++;
      }
    }
    else if(r == 6) ans += 2*(pow_(2, d - 1) + g(d - 1));
    else if(r == 7){
      for(int i = 7 * pow_(10, d - 1); i <= n; i++){
	s = 0; t = 0; f = 0; z = 0;
	temp = i;
	while(temp > 0){
	  if(temp % 10 == 3) t++;
	  else if(temp % 10 == 5) f++;
	  else if(temp % 10 == 7) s++;
	  else {z++; break;}
	}
	if(s*t*f > 0 && z == 0)ans++;
      }
    }
    else ans += g(d);
    printf("%d\n", ans);
  }
	
    
  return 0;
}