【CodeForces - 244B】Undoubtedly Lucky Numbers （dfs打表 + 二分）

题干：

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

Output

Input

Output

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：

输入一个n，定义Undoubtedly Lucky 数字是仅由小于等于两种数字组成的。问你有多少个小于等于n的是Undoubtedly Lucky 的数字。

解题报告：

写法很多啊这题也可以直接用set去重，，，dfs别写萎了就行了。。然后注意这题二分用upper_bound不能用lowerbound。。（因为要分很多种情况，，比如包含这个点或者不包含这个点）

AC代码：

#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 ;}

总结：

至于为什么main函数的内层for循环为什么要从i+1开始，这是因为我们现在是枚举的那两个数字，这样会保证没有重复啊，不然肯定会多搜一倍的东西。。

【CodeForces - 244B】Undoubtedly Lucky Numbers （dfs打表 + 二分）相关推荐

D - Undoubtedly Lucky Numbers CodeForces - 244B（数论）
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose deci ...
CodeForces - 1547F Array Stabilization (GCD version)(ST表+二分)
题目链接:点击查看题目大意:给出一个长度为 nnn 的数组 aaa,下标从 000 开始,每次操作分为两个步骤: 构建出数组 bbb,有 bi=gcd(ai,a(i+1)modn)b_i=gcd(a ...
HDU 5756 ztr loves lucky numbers （dfs）（搜索）
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5676 题目大意:只有7和4且7和4数量相等的数字为幸运数字,给你数字n,你要找出大于等于n的最小的幸运 ...
HDU 5676 ztr loves lucky numbers
-亚信科技,巴卡斯(杭州),壹晨仟阳(杭州),英雄互娱(杭州) (包括2016级新生)除了校赛,还有什么途径可以申请加入ACM校队? ztr loves lucky numbers Time Li ...
*【CodeForces - 122D】Lucky Transformation（字符串问题，思维剪枝，优化，有坑，需注意的问题if的层次总结）
题干: Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decima ...
【CodeForces - 122B 】Lucky Substring （字符串，水题）
题干: Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decima ...
*【CodeForces - 122C 】Lucky Sum （bfs记录状态，二分查找，有坑）（或分块）
题干: Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decima ...
HDU 2563 统计问题 (DFS + 打表)
统计问题 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submi ...
Covering（dfs打表+高斯消元+矩阵快速幂）
传送门这是个递推题,可以看下维基的各种解释:传送门假设递推式是一个四元方程,至于为什么也不大清楚(可能是操场是4*n的缘故) 然后得到递推式f(n)=f(n-1)+5*f(n-2)+f(n-3)- ...

【CodeForces - 244B】Undoubtedly Lucky Numbers （dfs打表 + 二分）

【CodeForces - 244B】Undoubtedly Lucky Numbers （dfs打表 + 二分）相关推荐

最新文章

热门文章