ZROI#1000

第一印象:啊,数位\(DP\),第一题这么清爽吗(内心\(:mmp\)).不过,这应该可以数位\(DP\)吧...不知道不知道,没想.
冷静思考\(\times 1\):对于线性的好像可以前缀和,因为这玩意儿非常稀疏,判定合法的\(log\)完全可以不管.

冷静思考\(\times 2\):抛弃什么数位\(DP\),这题是不是枚举位数+枚举与其他位不相同的数是啥+枚举它在哪+枚举其他位是啥就行了?
算一算复杂度\(:\Theta(17^2\times 9^2)\).哎呀,过了啊.写写写:
\(Code:\)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

int l , r , ans ;

inline bool check (int x , int cmp) {
    int cnt = 0 ;
    while  ( x ) ++ cnt , x /= 10 ;
    return cnt == cmp ;
}

signed main (int argc , char * argv[]) {
    l = rint () ; r = rint () ;
    rep ( i , 1 , 17 )
        rep ( b , 1 , i )
            rep ( j , 0 , 9 )
                rep ( k , 0 , 9 ) {
                    if ( j == k ) continue ; int res = 0 ;
                    rep ( w , 1 , b - 1 ) res = res * 10 + k ;
                    res = res * 10 + j ;
                    rep ( w , b + 1 , i ) res = res * 10 + k ;
                    if ( ! check ( res , i ) ) continue ;
                    if ( res >= l && res <= r ) ++ ans ;
                }
    printf ("%lld\n" , ans ) ;
    return 0 ;
}