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: Time Limit Exceeded)

#include<stdio.h>


/*int str_length(const char str[]) {

	int len=0;
	int i=0;
	while(str[i]) {
		len++;
		i++;
	}
	
	return(len);
}*/
/*整数の桁数を返す*/
int N_length(int i){
	int j=0;
	while(i>0) {
		i=i/10;
		j++;
	}
	return(j);
}


int main(void)
{
    long int N_in;
	
	scanf("%d", &N_in);
	int Num=0;
for(long int j=1; j<=N_in;j++) {
	int str[10]={0};
	/*printf("Now j=%d\n", j);*/
	/*printf("桁数は%dです\n", N_length(j));*/
	/*printf("桁数は%dです。",N_length(N_in));*/
	int temp=j;
	int next=0;
	for(int i=0; i<N_length(j);i++){
		str[i]=temp%10;
		if(str[i]!=3&&str[i]!=5&&str[i]!=7){
		    next=1;
		}
		
		temp=temp/10;
		/*printf("str[%d]=%d\n", i, str[i]);*/
		/*printf("temp=%d\n", temp);*/
		if(next==1) break;
	}
	
	
	if(next==0) {
	    int answer=1;
		int N_3=0;
		int N_5=0; 
		int N_7=0;
		int N_e=0;
		for(int i=0; i<N_length(j);i++){
		
			if(str[i]==7 ) { 
		    	N_7++;}
			else if(str[i]==5) {
			     	N_5++;}
			 	 else if(str[i]==3) {N_3++;}
			          else {N_e++;}
			/*printf("false str[%d]=%d\n", i, str[i]);*/
	    
			if(answer==0) break;
	}
	if(N_7>0&& N_5>0&&N_3>0&&N_e==0) {Num++;
		/*printf("Now j=%d",j);*/}
	
	
    }
}
	printf("%d", Num);
	return(0);
}