Codeforces-535B Tavas and SaDDas

Yvetteweiss

于 2017-05-06 20:01:57 发布

阅读量520

点赞数

分类专栏： ACM 文章标签： Codeforces

本文链接：https://blog.youkuaiyun.com/yvetteweiss/article/details/71289963

版权

ACM 专栏收录该内容

24 篇文章

订阅专栏

本文介绍了一种通过二分查找树的方法来确定给定幸运数在其所有同类数中位置的算法。幸运数是指十进制表示中仅包含4和7的正整数。通过输入一个幸运数n，算法能快速找出其在所有递增排列幸运数中的1基索引。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The problem is:
You are given a lucky number n. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what’s the 1-based index of n?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.

Input
The first and only line of input contains a lucky number n (1 ≤ n ≤ 109).

Output
Print the index of n among all lucky numbers.

Example
Input
4
Output
1

Input
7
Output
2

Input
77
Output
6

分析：乍一看好像很复杂，其实就是一棵二分查找树。每次判断最高位的数字：4—-cnt=2*cnt+1；7—-cnt=2*cnt+2。

Source：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
    char s[15];
    scanf("%s",s);
    int cnt=0;
    for(int i=0;i<strlen(s);i++)
    {
        if(s[i]=='4')
            cnt=2*cnt+1;
        else
            cnt=2*cnt+2;
    }
    printf("%d\n",cnt);
    return 0;
}