OJ Board

Code: Select all

void ProcessNumber(int n)
{
int k = 0, temp = n;
double  min = 0, max = 0;
		
		while (temp > 0) {
			temp /= 10;
			++k;
		}
		
		k+= 1;
		
		for (k; k < INT_MAX; ++k) {

			min = (log10(n) + k) / log10(2);
			max = (log10(n + 1) + k) / log10(2);
 /* this is the minium and maxmim range for E  but how can i choose a  particular E */
                         

		}
		
}

Code: Select all

#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<list>
#include<queue>
#include<cctype>
#include<stack>
#include<map>
#include<set>
using namespace std;
int n_digits(int n)
{
	int i=0;
	while(n>=10)
	{
		n/=10;
		i++;
	}
return i;
}
int main()
{
	
	unsigned int  n;
	long double lg=log((long double)10.0)/log((long double)2.0),lg2=log((long double)2.0);
	while(cin>>n)
	{
		int i=n_digits(n);
		for(long double j=i+2.0;;j++)
		{
			if(ceil(log((long double)n)/lg2+j*lg)==floor(log((long double)n+1.0)/lg2+j*lg))
			{
				printf("%.0Lf\n",ceil(log((long double)n)/lg2+j*lg));
				break;
			}
		}	
	}	
	return 0;
}

Code: Select all

#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<list>
#include<queue>
#include<cctype>
#include<stack>
#include<map>
#include<set>
using namespace std;
long double n_digits(long double n)
{
	long double i=0.0;
	while(n>=10.0)
	{
		n/=10.0;
		i+=1.0;
	}
return i;
}
int main()
{
	
	long double  n;
	while(cin>>n)
	{
		long double i=n_digits(n);
		for(long double j=i+2.0;;j+=1.0)
		{
			long double arg1= n;
			long double arg2=n+1.0;
			long double log2=log10(2.0);
			if(ceil((log10(arg1)+j)/log2)==floor((log10(arg2)+j)/log2))
			{
				printf("%.0Lf\n",ceil((log10(arg1)+j)/log2));
				break;
			}
		}	
	}	
	return 0;
}

Code: Select all

#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<list>
#include<queue>
#include<cctype>
#include<stack>
#include<map>
#include<set>
using namespace std;


int main()
{
	
	long double  n;
	while(cin>>n)
	{
		long double i=floor(log10(n));
		long double log2=log10((long double)2.0);
		for(long double j=i+2.0;;j+=1.0)
		{
			long double arg1= n;
			long double arg2=n+1.0;
			
			
			if(ceil((log10(arg1)+j)/log2)==floor((log10(arg2)+j)/log2))
			{
				printf("%.0Lf\n",ceil((log10(arg1)+j)/log2));
				break;
			}
		}	
	}	
	return 0;
}