CodeForces - 244B Undoubtedly Lucky Numbers（STL+思维）
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.
- 题目大意:
一个数中只含有两个或两个以下不同的数字就说这个数是幸运数
输入一个数n ，问小于等于n的数中有多少个幸运数 - 解题思路：
因为一共有0-9 九个数字 和0-9九个数字的组合 所以我们枚举出所有x,y所能够组成的数，然后看看<=n就加入set ,最后输出set内数的个数就是答案了。 - 代码如下：
#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;
}