CodeForces - 244B Undoubtedly Lucky Numbers（STL+思維）

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.

Examples
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.

• 題目大意:
  一個數中只含有兩個或兩個以下不同的數字就說這個數是幸運數
  輸入一個數n ，問小於等於n的數中有多少個幸運數
• 解題思路：
  因為一共有0-9 九個數字 和0-9九個數字的組合 所以我們枚舉出所有x,y所能夠組成的數，然後看看<=n就加入set ,最後輸出set內數的個數就是答案了。
• 程式碼如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#define ll long long
using namespace std;
set<ll>s;
ll n;
void find(ll x,ll y,ll num)
{
	s.insert(num);
	ll xx=x+num*10;
	ll yy=y+num*10;
	if(xx<=n&&xx!=0)
		find(x,y,xx);
	if(yy<=n&&yy!=0)
		find(x,y,yy);
	return ;
}

int main()
{
	cin>>n;
	s.clear();
	for(int i=1;i<=9;i++)
		for(int j=0;j<=9;j++)
			find(i,j,0);
	cout<<s.size()-1<<endl;
	return 0;
}