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: Accepted)

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

char univ[10000000][20] = {};
char arr[20] = {};
long univcnt = 0;

long conv(char c){
  switch(c){
    case '0':
      return 0;
    case '1':
      return 3;
    case '2':
      return 5;
    case '3':
      return 7;
  }
}
      
void make_pat(int floor, char num,int invfloor){
  if(floor == 1){
    arr[invfloor] = num;
    strcpy(univ[univcnt],arr);
    univcnt++;
  }else{
    arr[invfloor] = num;
    make_pat(floor-1,'0',invfloor+1);
    make_pat(floor-1,'1',invfloor+1);
    make_pat(floor-1,'2',invfloor+1);
    make_pat(floor-1,'3',invfloor+1);
  }
}
    
    
  

long power_ (long a,long n){
  long result=1;
  long i = 0;
  if(n == 0){
    return 1;
  }else{
    for(i=0;i<n;i++){
      result *= a;
    }
    return result;
  }
}
    
int main(){
  long n = 0;
  long ndeg = 0;
  long i = 0;
  long j = 0;
  long ok = 0;
  long check = 0;
  long num = 0;
  long cnt = 0;
  int threecheck = 0;
  int fivecheck = 0;
  int sevencheck = 0;
  scanf("%ld",&n);
  long a[10] = {};
  make_pat(10,'0',0);
  make_pat(10,'1',0);
  make_pat(10,'2',0);
  make_pat(10,'3',0);
  for(i=0;i<univcnt;i++){
    ok = 1;
    threecheck = 0;
    fivecheck = 0;
    sevencheck = 0;
    check = 0;
    
    for(j=0;j<10;j++){
      if(univ[i][j] == '1'){
        threecheck = 1;
        check = 1;
      }else if(univ[i][j] == '2'){
        fivecheck = 1;
        check = 1;
      }else if(univ[i][j] == '3'){
        sevencheck = 1;
        check = 1;
      }else{
        if(check){
          ok = 0;
        }
      }
    }
    
    if(ok&&threecheck&&fivecheck&&sevencheck){
      //printf("%s/",univ[i]);
      num = 0;
      for(j=0;j<10;j++){
        num += conv(univ[i][9-j])*power_(10,j);
      }
      //printf("%ld/",num);
      if(num <= n){
        cnt++;
      }
    }
  }
  printf("%ld",cnt);
  return 0;
}