Codeforces Round #150 (Div. 2) B dfs

http://codeforces.com/contest/244

题意：

给一个数n (1 ≤ n ≤ 10⁹) ，然后求小于等于n的数，该数并且满足只有两个十进制数(0-9)组成的个数；

思路：

当时就是一门心思推公式，结果还是没找出规律。赛后想了想推个毛公式啊直接暴力枚举n的长度，然后枚举0到9两辆组合时间复杂度为O(10*10*10*2^10) = O(10^6)啊。

哎只怪自己没有想出来吧。这里还要注意当枚举长度为10时可能会出现超数据类型的要用__int64

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 1073741824
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 150
#define N 1007
using namespace std;
//freopen("data.in","r",stdin);

char num[15];
int mk[15];
int no;
set<ll>st;

void dfs(int L,int a,int b,int pos){
    if (pos == L + 1){
        ll sum = 0;
        for (int i = 1; i <= L; ++i){
            sum = sum*10;
            if (mk[i] == 1) sum += a;
            else sum += b;
        }
        if (sum <= no){
            st.insert(sum);
        }
        return ;
    }
    mk[pos] = 1;
    dfs(L,a,b,pos + 1);
    mk[pos] = 0;
    dfs(L,a,b,pos + 1);
}
int main(){
    //freopen("din.txt","r",stdin);
    int i,j,k;
    while (~scanf("%s",num)){
        int len = strlen(num);
        no = 0;
        for (i = 0; i < len; ++i){
            no = no*10 + (num[i] - '0');
        }
        //cout<<no<<"  " <<len<<endl;
        if (no >= 1 && no <= 99){
            printf("%d\n",no);
        }
        else{
            st.clear();
            for (i = 2; i <= len; ++i){
                for (j = 0; j <= 9; ++j){
                    for (k = 0; k <= 9; ++k){
                        CL(mk,0);
                        dfs(i,j,k,1);
                    }
                }
            }
            cout<<st.size() - 1<<endl;
        }
    }


}