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>
int n;
int ten[10]={1,10,100,1000,10000,100000,1000000,10000000,100000000};
int aaa[4]={3,5,7};
int result=0;
int check(int num);
int xxx(int i,int j){
	for(int k=0;k<3;k++)
	{
		if(check(10*j+aaa[k])==1&&10*j+aaa[k]<=n)
		{
			result++;
		}
		if(10*j+aaa[k]>n)
		{
			return 0;
		}
		if(i<9)
		{
			xxx(i+1,10*j+aaa[k]);
			
		}
	}
	return 0;
}

int check(int num){
	int three=0,five=0,seven=0;
	for(int i=0;i<9;i++)
	{
		if((num/ten[i])%10==3)three=1;
		if((num/ten[i])%10==5)five=1;
		if((num/ten[i])%10==7)seven=1;
	}
	
	if(seven==1&&five==1&&three==1)
	{
		return 1;
	}
	return 0;
	}


int main(){
	
	
	scanf("%d",&n);
	
	xxx(0,0);
	printf("%d",result);
	}