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

int foo(char * c)
{
  int rc = 0;
  if( *c == ' ' ){
    *c = '3';
  } else if( *c == '3' ){
    *c = '5';
  } else if( *c == '5' ){
    *c = '7';
  } else if( *c == '7' ){
    *c = '3';
    rc = 1;
  }
  return rc;
}

int main()
{
  char orig[12],digi[12],chk[3];
  int n,i,j,cnt;
  scanf("%d",&n);
  sprintf(orig,"%10d",n);
  /* printf("%s\n",orig); */
  for(i=0;i<10;i++){ digi[i] = ' '; }
  digi[10] = (char)0;
  cnt = 0;
  while(1){
    if( foo(&(digi[9])) ){
      for(j=8;j>=0;j--){
        if(!foo(&(digi[j]))) break;
      }
    }
    if( digi[0] != ' ' ) break;
    memset(chk,0,sizeof(chk));
    for(i=0;i<10;i++){
      if(digi[i]=='3') chk[0] ++;
      if(digi[i]=='5') chk[1] ++;
      if(digi[i]=='7') chk[2] ++;
    }
    if( chk[0]==0 || chk[1]==0 || chk[2]==0 ) continue;
    sscanf(digi,"%d",&i);
    if( i>n ) break;
    cnt++;
  }
  printf("%d\n",cnt);
  
  return EXIT_SUCCESS;
}