Problem - 244B - Codeforces STL 中set的用法

http://codeforces.com/problemset/problem/244/B

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：用x和y组成不大于n的整数有多少个(0 ≤ x,  y ≤ 9，1 ≤ n ≤ 10^9）。

——>>枚举x, y，将其所有组成的数放入set，最后看set里有多少个元素就好。

set中的值都是特定的，而且只出现一次.

定义一个元素为整数的集合a，

可以用 set<int> a;

 基本操作： 

对集合a中元素的有 插入元素：a.insert(1); 

删除元素（如果存在）：a.erase(1); 
判断元素是否属于集合：if (a.find(1) != a.end()) ... 

返回集合元素的个数：a.size() 

将集合清为空集：a.clear()

#include <stdio.h>#include <set>using namespace std;int n;set<long long>se;void find(int x,int y,long long cur){    long long getx=cur*10+x;    long long gety=cur*10+y;    if(getx&&getx<=n)    {        se.insert(getx);        find(x,y,getx);    }    if(gety<=n)    {        se.insert(gety);        find(x,y,gety);    }}int main(){    while(~scanf("%d",&n))    {        se.clear();        for(int x=0;x<9;x++)           for(int y=x+1;y<=9;y++)               find(x,y,0);        printf("%d\n",se.size());    }    return 0;}