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

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

//S[0]から始まる文字列の長さを求める関数
int string_len(char* str){
    int len=0;
    while(str[len]!=0){
        len++;
    }
    return len;
}

//753数なら1,そうでないなら0
int judge_753(int N){
    //int型Nを、文字列char配列に変換
    char s[10]={'\0'};
    snprintf(s,10,"%d",N);

    //Nの桁数
    int digit=string_len(s);

    int count=0;
    for(int i=0;i<digit;i++){
        if(s[i]=='3'){
            count++;
            break;
        }
    }
    for(int i=0;i<digit;i++){
        if(s[i]=='5'){
            count++;
            break;
        }
    }
    for(int i=0;i<digit;i++){
        if(s[i]=='7'){
            count++;
            break;
        }
    }

    if(count==3) return 1;
    else return 0;
}

void count_753(int s,int N, int ans,int a, int b, int c){
    if(s>N) return;

    if(a+b+c==3) ans++;

    count_753((s*10+3),N,ans,1,b,c);
    count_753((s*10+5),N,ans,a,1,c);
    count_753((s*10+7),N,ans,a,b,1);
}

int main(void){
    int N;
    scanf("%d",&N);

    int ans=0;
    count_753(0,N,ans,0,0,0);

    printf("%d\n",ans);

    return 0;
}