Warm up 13 [D] Different Digits

 

Different Digits

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 45, Accepted users: 45

Problem 12751 : No special judgement

 

Problem description

The inhabitants of Nlogonia are very superstitious. One of their beliefs is that street house numbers that have a repeated digit bring bad luck for the residents. Therefore, they would never live in a house which has a street number like 838 or 1004.
The Queen of Nlogonia ordered a new seaside avenue to be built, and wants to assign to the new houses only numbers without repeated digits, to avoid discomfort among her subjects. You have been appointed by Her Majesty to write a program that, given two integers N and M, determines the maximum number of houses that can be assigned street numbers between N and M, inclusive, that do not have repeated digits.

Input

Each test case is described using one line. The line contains two integers N and M, as described above (1 ≤ N ≤ M ≤ 5000).

Output

For each test case output a line with an integer representing the number of street house numbers between N and M, inclusive, with no repeated digits.

Sample Input

87 104
989 1022
22 25
1234 1234

Sample Output

14
0
3
1

Problem Source

Latin American 2012

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 1000005
#define mod 1000000007
#define ll long long
#define INF 0x7fffffff
#define FF(i,n)for(int i=0;i<n;i++)
int n, m;
int fun(int x){
    int a[5];
    int i = 0;
    while (x){
        a[i++] = x % 10;
        x /= 10;
    }
    for (int j = 0; j < i;j++)
    for (int k = j+1; k < i;k++)
    if (a[k] == a[j])return 0;
    return 1;
}
int main(){
    int cas = 1;
    int t;
    /*scanf("%d", &t);
    while (t--){
        
    }*/
    while (~scanf("%d%d", &n,&m)){
        t = 0;
        for (int i = n; i <= m; i++){
            if (fun(i))t++;
        }
        printf("%d\n", t);
    }
    return 0;
}