codeforces 224B Undoubtedly Lucky Numbers

题目链接：http://codeforces.com/problemset/problem/244/B

解题思路：DFS 搜索

/**************************************************************************
user_id:                                                    SCNU20102200088
problem_id:                                                 codeforces 244B
problem_name:                                     Undoubtedly Lucky Numbers
**************************************************************************/

#include <algorithm>
#include <iostream>
#include <iterator>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;

//线段树
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

//手工扩展栈
#pragma comment(linker,"/STACK:102400000,102400000")

const double EPS=1e-9;
const double PI=acos(-1.0);
const double E=2.7182818284590452353602874713526;  //自然对数底数
const double R=0.5772156649015328606065120900824;  //欧拉常数:(1+1/2+...+1/n)-ln(n)

const int x4[]={-1,0,1,0};
const int y4[]={0,1,0,-1};
const int x8[]={-1,-1,0,1,1,1,0,-1};
const int y8[]={0,1,1,1,0,-1,-1,-1};

typedef long long LL;

typedef int T;
T max(T a,T b){ return a>b? a:b; }
T min(T a,T b){ return a<b? a:b; }
T gcd(T a,T b){ return b==0? a:gcd(b,a%b); }
T lcm(T a,T b){ return a/gcd(a,b)*b; }

///////////////////////////////////////////////////////////////////////////
//Add Code:
set<LL> s;

void DFS(int n,int i,int j,LL res,int cnt){
    if(res>n || (res==0 && cnt>0)) return ;
    if(res>0) s.insert(res);
    DFS(n,i,j,10*res+i,cnt+1);
    DFS(n,i,j,10*res+j,cnt+1);
}
///////////////////////////////////////////////////////////////////////////

int main(){
    std::ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    ///////////////////////////////////////////////////////////////////////
    //Add Code:
    int n,i,j;
    scanf("%d",&n);
    for(i=0;i<9;i++){
        for(j=i+1;j<=9;j++) DFS(n,i,j,0,0);
    }
    printf("%d
",s.size());
    ///////////////////////////////////////////////////////////////////////
    return 0;
}

/**************************************************************************
Testcase:
Input:
10
123
Output:
10
113
**************************************************************************/