Codeforces Round #150 (Div. 2) B. Undoubtedly Lucky Numbers

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B. Undoubtedly Lucky Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 are lucky.

Let's call a positive integer a undoubtedly lucky, if there are such digits x and y (0 ≤ x, y ≤ 9), that the decimal representation of number a (without leading zeroes) contains only digits x and y.

Polycarpus has integer n. He wants to know how many positive integers that do not exceed n, are undoubtedly lucky. Help him, count this number.

Input

The first line contains a single integer n (1 ≤ n ≤ 109) — Polycarpus's number.

Output

Print a single integer that says, how many positive integers that do not exceed n are undoubtedly lucky.

Sample test(s)

input

10

output

10

input

123

output

113

Note

In the first test sample all numbers that do not exceed 10 are undoubtedly lucky.

In the second sample numbers 102, 103, 104, 105, 106, 107, 108, 109, 120, 123 are not undoubtedly lucky.

题意：如果一个数只由两个或一个数字组成，那么说明它是一个Undoubtedly Lucky Numbers，输入一个n，问从1到n一共有多少这样的数？      

起初考虑的是用set，后来想了想发现只用set不可以，后来发现这种数是有规律的，于是变成了一个简单的深搜题。

代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<set>#include<algorithm>using namespace std;set<int>ss;set<int>::iterator iter; int x,y; long long n,i;void dfs(long long sa,int ans) {  if(sa>n||ans>10) return ;   ss.insert(sa);   dfs(sa*10+x,ans+1);   dfs(sa*10+y,ans+1); } int main() {   cin>>n;   for(x=0;x<10;x++)    for(y=x+1;y<10;y++)      dfs(0,0);/*for(iter=ss.begin();iter!=ss.end();iter++)   {       cout<<*iter<<"&&";   }*/     printf("%d\n",ss.size()-1);     return 0; }