Critical – Live with passion

1314

Asif NaeemApril 28, 2015

#include <iostream>
#include <cstdio>
#include <sstream>
#include <string>

using namespace std;

int n;
string str;

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		cin>>n;
		cin>>str;
		int i=0,j=1,k;

		while(i<n && j<n)
		{
			for(k=0;k<n;k++)
			{
				if(str[(k+i)%n]!=str[(k+j)%n])
				break;
			}
			if(k==n)
			break;
			if(str[(k+i)%n]>str[(k+j)%n])
				i=i+k+1;
			else if(str[(k+i)%n]<str[(k+j)%n])
				j=j+k+1;
			if(i==j)
			j+=1;
		}
		cout<<((i>j)?j:i)<<"\n";
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#include <sstream>

#include <string>

using namespace std;

int n;

string str;

int main()

{

int t;

cin>>t;

while(t--)

{

cin>>n;

cin>>str;

int i=0,j=1,k;

while(i<n && j<n)

{

for(k=0;k<n;k++)

{

if(str[(k+i)%n]!=str[(k+j)%n])

break;

}

if(k==n)

break;

if(str[(k+i)%n]>str[(k+j)%n])

i=i+k+1;

else if(str[(k+i)%n]<str[(k+j)%n])

j=j+k+1;

if(i==j)

j+=1;

}

cout<<((i>j)?j:i)<<"\n";

}

return 0;

}

Theory: We know that, From 0-9, each digit occurs once From 0-99, each digit occurs 20 times (10x in position 1 and 10x in position 2) From 0-999, each digit occurs 300 times (100x in P1, 100x in P2, 100x in P3) If the range is from 0 to a power of 10 then occurrence

Continue Reading “12241” →

12517

Asif NaeemNovember 18, 2014

Theory: from 0 to 99 there are 2*10^1=20 for each 0 to 9 digits. from 0 to 999 there are 3*10^2=300 for each 0 to 9 digits. so the foemulae is for 0 to n number of 9’s n*10^(n-1) for each 0 to 9 digits. And so, for upto 445 we can find total number

Continue Reading “12517” →

1016

Asif NaeemOctober 28, 2014

Theory: The key to figuring out this problem was to recognize that you needed to resolve cycles in the data – if you follow an item to the item that’s in its spot, and then the one that’s in the next one’s spot, until you reach the original item again, you’ll have a cycle of

Continue Reading “1016” →

306

Asif NaeemSeptember 20, 2014

#include <iostream>
#include <map>
#include <vector>
#include <string>
#include <stdio.h>

/*

10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0

Sample Output

BolHeol  b
C RCE

 */
using namespace std;

int arr[202];
vector<char> c_arr[202];
char s[202];


int cycle(int pos)
{
	int org=pos;
	int init=arr[pos];
	int cnt=1;
	c_arr[pos-1].clear();
	c_arr[org-1].push_back(s[arr[pos]]);

	pos=arr[pos];

	while(init!=arr[pos])
	{
		c_arr[org-1].push_back(s[arr[pos]]);
		pos=arr[pos];
		cnt++;
	}

	return cnt;
}

int main()
{
	int n,k,len;
	string str;
	int pos_of_cycle;

	while(cin>>n)
	{
		if(n==0)
		break;
		for(int i=1;i<=n;i++)
		{
			cin>>arr[i];
		}
		while(cin>>k)
		{
			getchar();
			if(k==0)
			break;

			getline(cin,str);

			for(int i=0;i<str.size();i++)
			{
				s[i+1]=str[i];
			}
			for(int i=(str.size()+1);i<=n;i++)
				s[i]=' ';
			for(int pos=1;pos<=n;pos++)
			{
				len = cycle(pos);
				pos_of_cycle=len-1-(k%len);
				cout<<c_arr[pos-1][pos_of_cycle];

			}
			cout<<"\n";
		}
		cout<<"\n";

	}

	return 0;
}

#include <iostream>

#include <map>

#include <vector>

#include <string>

#include <stdio.h>

4 5 3 7 2 8 1 6 10 9

1 Hello Bob

1995 CERC

Sample Output

BolHeol b

C RCE

using namespace std;

int arr[202];

vector<char> c_arr[202];

char s[202];

int cycle(int pos)

{

int org=pos;

int init=arr[pos];

int cnt=1;

c_arr[pos-1].clear();

c_arr[org-1].push_back(s[arr[pos]]);

pos=arr[pos];

while(init!=arr[pos])

{

c_arr[org-1].push_back(s[arr[pos]]);

pos=arr[pos];

cnt++;

}

return cnt;

}

int main()

{

int n,k,len;

string str;

int pos_of_cycle;

while(cin>>n)

{

if(n==0)

break;

for(int i=1;i<=n;i++)

{

cin>>arr[i];

}

while(cin>>k)

{

getchar();

if(k==0)

break;

getline(cin,str);

for(int i=0;i<str.size();i++)

{

s[i+1]=str[i];

}

for(int i=(str.size()+1);i<=n;i++)

s[i]=' ';

for(int pos=1;pos<=n;pos++)

{

len = cycle(pos);

pos_of_cycle=len-1-(k%len);

cout<<c_arr[pos-1][pos_of_cycle];

}

cout<<"\n";

}

cout<<"\n";

}

return 0;

}

12712

Asif NaeemSeptember 1, 2014

#include <iostream>

using namespace std;

int main()
{
	int t,c=0;
	int MX;
	int L,M,N;
	unsigned long long sum,tmp;

	cin>>t;

	while(t--)
	{
		cin>>L>>M>>N;
		c++;
		MX=L*L;
		tmp=1;
		for(int i=M-1;i>=0;i--)
		{
			tmp=tmp*(MX-i);
			tmp%=10000000000007;
		}

		sum=0;

		for(int j=1;j<=N-M+1;j++)
		{
			sum+=tmp;
			sum%=10000000000007;
			tmp*=(MX-M+1-j);
			tmp%=10000000000007;
		}

		cout<<"Case "<<c<<": ";
		cout<<sum<<"\n";
	}

	return 0;
}

#include <iostream>

using namespace std;

int main()

{

int t,c=0;

int MX;

int L,M,N;

unsigned long long sum,tmp;

cin>>t;

while(t--)

{

cin>>L>>M>>N;

c++;

MX=L*L;

tmp=1;

for(int i=M-1;i>=0;i--)

{

tmp=tmp*(MX-i);

tmp%=10000000000007;

}

sum=0;

for(int j=1;j<=N-M+1;j++)

{

sum+=tmp;

sum%=10000000000007;

tmp*=(MX-M+1-j);

tmp%=10000000000007;

}

cout<<"Case "<<c<<": ";

cout<<sum<<"\n";

}

return 0;

}

12716

Asif NaeemAugust 31, 2014

#include <iostream>
using namespace std;

int a[30000005];
int main() {

   int t,n,c=0;
   for(int i=1;i<30000005;i++)
   {
      for(int j=i<<1;i+j<30000005;j+=i)
      {
        if(((i+j)&j)==j)
        a[i+j]++;
      }
   }

   for(int i=1;i<30000005;i++)
   a[i]+=a[i-1];
   cin>>t;

   while(t--)
   {
      cin>>n;
      c++;
      cout<<"Case "<<c<<": "<<a[n]<<"\n";
   }

   return 0;
}

#include <iostream>

using namespace std;

int a[30000005];

int main() {

int t,n,c=0;

for(int i=1;i<30000005;i++)

{

for(int j=i<<1;i+j<30000005;j+=i)

{

if(((i+j)&j)==j)

a[i+j]++;

}

for(int i=1;i<30000005;i++)

a[i]+=a[i-1];

cin>>t;

while(t--)

{

cin>>n;

c++;

cout<<"Case "<<c<<": "<<a[n]<<"\n";

}

return 0;

}

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