題干：

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?nare 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：

輸入一個n，定義Undoubtedly Lucky 數字是僅由小于等于兩種數字組成的。問你有多少個 小于等于n的 是Undoubtedly Lucky 的數字。

解題報告：

? ?寫法很多啊這題也可以直接用set去重，，，dfs別寫萎了就行了。。然后注意這題二分用upper_bound不能用lowerbound。。（因為要分很多種情況，，比如包含這個點或者不包含這個點）

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll up; int tot; ll a[MAX*100]; void dfs(ll cur,ll wei,ll i,ll j) {if(wei > 10 /*|| cur > up*/) return ;a[++tot] = cur; // if(cur == 101) { // putchar('a'); // }dfs(cur*10+i,wei+1,i,j);dfs(cur*10+j,wei+1,i,j); } int main() {cin>>up;for(int i = 0; i<=9; i++) {for(int j = i+1; j<=9; j++) {dfs(0,0,i,j);}}sort(a+1,a+tot+1);int len = unique(a+1,a+tot+1) - a-1; // printf("len = %d\n",len);for(int i = 1; i<=len; i++) printf("%lld ",a[i]); // puts("");printf("%d\n",upper_bound(a+1,a+len+1,up) - a - 1 -1);return 0 ;}

總結：

? 至于為什么main函數的內層for循環為什么要從i+1開始，這是因為我們現在是枚舉的那兩個數字，這樣會保證沒有重復啊，不然肯定會多搜一倍的東西。。

總結

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