codeforce 9C 找规律的思维题

最新推荐文章于 2023-06-11 22:42:16 发布

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最新推荐文章于 2023-06-11 22:42:16 发布

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本文介绍了一个算法，用于解决给定范围内能被Hexadecimal内存系统成功存储的十进制整数数量的问题。通过将输入的整数转换为字符串，并逐位检查其二进制表示是否仅包含0和1来实现。

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http://vjudge.net/contest/view.action?cid=47681#problem/C

Description

One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.

But his plan failed. The reason for this was very simple: Hexadecimal didn't perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.

Input

Input data contains the only number n (1 ≤ n ≤ 10⁹).

Output

Output the only number — answer to the problem.

Sample Input

Input

Output

Hint

For n = 10 the answer includes numbers 1 and 10.

用a[ i],数组来表示长度为i的10 ^i,说含有的可取的数。把输入的数字当做字符串来处理，如果一个位上的数大于1，就加上a[1~i]，若为1，则就加上a[i]，0则不加。

注意其实数据范围是1~10^10，，，题干描述不准确，我被骗wa了一次==

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
char a[10];
int sum[11]={0,1,2,4,8,16,32,64,128,256,512};
int main()
{
    while(~scanf("%s",a+1))
    {
        int n=strlen(a+1);
        int sum1=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]>'1')
            {
                for(int j=1;j<=n-i+1;j++)
                    sum1+=sum[j];
                break;
            }
            if(a[i]=='1')
                sum1+=sum[n-i+1];
        }
        printf("%d\n",sum1);
    }
    return 0;
}