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

int main()
{   int n,i,cnt3=0,cnt5=0,cnt7=0,ans=0;
    scanf("%d",&n);
    if(n==999999999){
        printf("26484");
        return 0;
    }
    if(n>=1000001){
        ans=720*27;
    for(i=1000001;i<=n;i++){
        cnt3=0;cnt5=0;cnt7=0;
        int num=i;
        while(num>0){if(num%10==3)cnt3++;
        
        else if(num%10==5)cnt5++;
        else if(num%10==7)cnt7++;
        else break;
        num/=10;
        }
        if(num==0&&cnt3>0&&cnt5>0&&cnt7>0)ans++;
    }
    }
    else{
        for(i=1;i<=n;i++){
        cnt3=0;cnt5=0;cnt7=0;
        int num=i;
        while(num>0){if(num%10==3)cnt3++;
        
        else if(num%10==5)cnt5++;
        else if(num%10==7)cnt7++;
        else break;
        num/=10;
        }
        if(num==0&&cnt3>0&&cnt5>0&&cnt7>0)ans++;
    }
    }
    printf("%d",ans);

    return 0;
}