9C - Hexadecimal's Numbers - CodeForces Solution


brute force implementation math *1200

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Python Code:

n = input()
l = ''
for i in n:
    if i=='0':
        l +='0'
    elif i=='1':
        l+='1'
    else:
        break 
l+='1'*(len(n)-len(l))
print(int(l,2))

C++ Code:

#include <bits/stdc++.h>
using namespace std;
#define  fast ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
const long int N=1e6;
 
 
int main()
{
	fast;
   string n;
   cin>>n;
   long long int som=0;
	vector<long long int>dp(n.length());
	dp[0]=1;
	for(long long int i(1);i<(n.length());i++){
		dp[i]=dp[i-1];
		dp[i]*=2;}
	for(long long int i(0);i<(n.length());i++){
		som+=dp[i];}
	bool test=true;
	for(long long int i(0);i<(n.length()-1);i++){
		if(n[i]>'1'){test=false;break;}
		if(n[i]=='0'){
			som-=dp[n.length()-1]/dp[i];}}
if(test!=false){
if(n[n.length()-1]=='0'){
	som-=1;}}
	cout<<som<<endl;
  
   
   
	
 
 
 
	
	
return 0;}


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