codeforces 9C (DFS）之 Hexadecimal's Numbers

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

10

Output

2

Hint

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

题意：给你一个数n，问你 ： 从1到n一共有多少个只由1,0组成的数？

分析：刚开始试了一下递推，发现不行，改用dfs了

个人感觉这一题有问题，循环一组测试数据之后，要打印累加的值才可以AC

AC代码如下：

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;int s[1000000];int a[1000000];int main(){    int ans;    int n;    a[1]=1;    for(int i=2;i<100;i++)        a[i]=a[i-1]+pow(2,i-1);    while(cin>>n)    {        ans=0;        memset(s,0,sizeof(s));        int num=0;        while(n!=0)        {            s[num]=n%10;            if(s[num]>=1) ans+=a[num];            if(s[num]>1) ans+=pow(2,num);            else if(s[num]==1) ans++;            n/=10;            num++;        }        cout<<ans<<endl;    }    return 0;}