Math – Live with passion

Karatsuba’s Fast Multiplication Algorithm

Asif NaeemJuly 25, 2017

Problem: Find multiplication result of two large numbers in less than O(n^2) time complexity.

#include <iostream>
#include <string>

/*
123456
34343
add=157799
sub=0@J113
*/
using namespace std;

void makeEqualString(string &a,string &b)
{
	if(a.size()==b.size())
	return;
	else
	{
		if(a.size()>b.size())
		{
			for(int i=1;i<=a.size()-b.size();i++)
				b='0'+b;
		}
		else
		{
			for(int i=1;i<=b.size()-a.size();i++)
				a='0'+a;
		}
	}
}

string addString(string a,string b)
{
	int idx1=a.size()-1;
	int idx2=b.size()-1;
	int carry=0;
	char c;
	string ans="";
	int tmp=0;
	while(idx1>=0 && idx2>=0)
	{
		tmp=(a[idx1]-'0')+(b[idx2]-'0')+carry;
		carry=tmp/10;
		tmp=tmp%10;
		c=tmp+'0';
		ans=c+ans;
		idx1-=1;
		idx2-=1;
	}
	if(idx1>=0)
	{
		while(idx1>=0)
		{
			tmp=(a[idx1]-'0')+carry;
			carry=tmp/10;
			tmp=tmp%10;
			c=tmp+'0';
			ans=c+ans;
			idx1-=1;
		}
	}
	
	if(idx2>=0)
	{
		while(idx2>=0)
		{
			tmp=(b[idx2]-'0')+carry;
			carry=tmp/10;
			tmp=tmp%10;
			c=tmp+'0';
			ans=c+ans;
			idx2-=1;
		}
	}
	if(carry)
	{
		c=carry+'0';
		ans=c+ans;
	}
	return ans;
}

string subtrachString(string a,string b)
{
	int idx1=a.size()-1;
	int idx2=b.size()-1;
	int carry=0;
	string ans="";
	int tmp=0,tmp2=0;
	char c;
	
	while(idx1>=0 && idx2>=0)
	{
		if((a[idx1]-'0')<((b[idx2]-'0')+carry))
		{
			tmp=(a[idx1]-'0')+10-(b[idx2]-'0')-carry;
			carry=1;
		}
		else
		{
			tmp=(a[idx1]-'0')-((b[idx2]-'0')+carry);
			carry=0;
		}
		c=tmp+'0';
		ans=c+ans;
		
		idx1-=1;
		idx2-=1;
	}
	
	if(idx1>=0)
	{
		while(idx1>=0)
		{
			if((a[idx1]-'0')<carry)
			{
				tmp=(a[idx1]-'0')+10-carry;
				carry=1;
			}
			else
			{
				tmp=(a[idx1]-'0')-carry;
				carry=0;
			}
			c=tmp+'0';
			ans=c+ans;
			idx1-=1;
		}
	}
	return ans;
}

string karatsuba(string a,string b)
{
	int res,x,y;
	string ans="";
	if(a.size()<=1 && b.size()<=1)
	{
		res=(a[0]-'0')*(b[0]-'0');
		if(res>=10)
		{
			x=res%10;
			res/=10;
			y=res%10;
			ans+=(y+'0');
			ans+=(x+'0');
		}
		else
		{
			x=res%10;
			ans+=(x+'0');
		}
		return ans;
	}
	makeEqualString(a,b);
	//cout<<a<<" "<<b<<endl;
	int _pow=a.size()/2;
	
	string xL=a.substr(0,_pow);
	string xH=a.substr(_pow,a.size());
	//cout<<"x="<<xL<<" "<<xH<<endl;
	
	string yL=b.substr(0,_pow);
	string yH=b.substr(_pow,b.size());
	//cout<<"y="<<yL<<" "<<yH<<endl;
	
	string e = karatsuba(xL,yL);
	
	string f = karatsuba(xH,yH);
	
	string g= addString(xL,xH);
	string h= addString(yL,yH);
	
	string gh = karatsuba(g,h);
	string subTrct = addString(e,f);
	
	string d = subtrachString(gh,subTrct);
	
	for(int i=1;i<=(a.size()-_pow)*2;i++)
		e=e+'0';
	for(int i=1;i<=(a.size()-_pow);i++)
		d=d+'0';
		
	ans = addString(e,d);
	ans = addString(ans,f);
	return ans;
}

int main()
{
	string a,b,ans;
	int t,i;
	cin>>t;
	while(t--)
	{
		cin>>a>>b;
		//cout<<addString(a,b)<<endl;
		//cout<<subtrachString(a,b)<<endl;
		ans=karatsuba(a,b);
		i=0;
		for(i=0;i<ans.size();i++)
		{
			if(ans[i]!='0')
			break;
		}
		//cout<<i<<" = i"<<endl;
		while(i!=ans.size())
			cout<<ans[i++];
		cout<<endl;
	}
	return 0;
}

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#include <iostream>

#include <string>

123456

34343

add=157799

sub=0@J113

using namespace std;

void makeEqualString(string &a,string &b)

{

if(a.size()==b.size())

return;

else

{

if(a.size()>b.size())

{

for(int i=1;i<=a.size()-b.size();i++)

b='0'+b;

}

else

{

for(int i=1;i<=b.size()-a.size();i++)

a='0'+a;

}

string addString(string a,string b)

{

int idx1=a.size()-1;

int idx2=b.size()-1;

int carry=0;

char c;

string ans="";

int tmp=0;

while(idx1>=0 && idx2>=0)

{

tmp=(a[idx1]-'0')+(b[idx2]-'0')+carry;

carry=tmp/10;

tmp=tmp%10;

c=tmp+'0';

ans=c+ans;

idx1-=1;

idx2-=1;

}

if(idx1>=0)

{

while(idx1>=0)

{

tmp=(a[idx1]-'0')+carry;

carry=tmp/10;

tmp=tmp%10;

c=tmp+'0';

ans=c+ans;

idx1-=1;

}

if(idx2>=0)

{

while(idx2>=0)

{

tmp=(b[idx2]-'0')+carry;

carry=tmp/10;

tmp=tmp%10;

c=tmp+'0';

ans=c+ans;

idx2-=1;

}

if(carry)

{

c=carry+'0';

ans=c+ans;

}

return ans;

}

string subtrachString(string a,string b)

{

int idx1=a.size()-1;

int idx2=b.size()-1;

int carry=0;

string ans="";

int tmp=0,tmp2=0;

char c;

while(idx1>=0 && idx2>=0)

{

if((a[idx1]-'0')<((b[idx2]-'0')+carry))

{

tmp=(a[idx1]-'0')+10-(b[idx2]-'0')-carry;

carry=1;

}

else

{

tmp=(a[idx1]-'0')-((b[idx2]-'0')+carry);

carry=0;

}

c=tmp+'0';

ans=c+ans;

idx1-=1;

idx2-=1;

}

if(idx1>=0)

{

while(idx1>=0)

{

if((a[idx1]-'0')<carry)

{

tmp=(a[idx1]-'0')+10-carry;

carry=1;

}

else

{

tmp=(a[idx1]-'0')-carry;

carry=0;

}

c=tmp+'0';

ans=c+ans;

idx1-=1;

}

return ans;

}

string karatsuba(string a,string b)

{

int res,x,y;

string ans="";

if(a.size()<=1 && b.size()<=1)

{

res=(a[0]-'0')*(b[0]-'0');

if(res>=10)

{

x=res%10;

res/=10;

y=res%10;

ans+=(y+'0');

ans+=(x+'0');

}

else

{

x=res%10;

ans+=(x+'0');

}

return ans;

}

makeEqualString(a,b);

//cout<<a<<" "<<b<<endl;

int _pow=a.size()/2;

string xL=a.substr(0,_pow);

string xH=a.substr(_pow,a.size());

//cout<<"x="<<xL<<" "<<xH<<endl;

string yL=b.substr(0,_pow);

string yH=b.substr(_pow,b.size());

//cout<<"y="<<yL<<" "<<yH<<endl;

string e = karatsuba(xL,yL);

string f = karatsuba(xH,yH);

string g= addString(xL,xH);

string h= addString(yL,yH);

string gh = karatsuba(g,h);

string subTrct = addString(e,f);

string d = subtrachString(gh,subTrct);

for(int i=1;i<=(a.size()-_pow)*2;i++)

e=e+'0';

for(int i=1;i<=(a.size()-_pow);i++)

d=d+'0';

ans = addString(e,d);

ans = addString(ans,f);

return ans;

}

int main()

{

string a,b,ans;

int t,i;

cin>>t;

while(t--)

{

cin>>a>>b;

//cout<<addString(a,b)<<endl;

//cout<<subtrachString(a,b)<<endl;

ans=karatsuba(a,b);

i=0;

for(i=0;i<ans.size();i++)

{

if(ans[i]!='0')

break;

}

//cout<<i<<" = i"<<endl;

while(i!=ans.size())

cout<<ans[i++];

cout<<endl;

}

return 0;

}

Steins GCD Algorithm

Asif NaeemSeptember 28, 2016

Algorithm to find GCD using Stein’s algorithm gcd(a,b) If both and b are 0, gcd is zero gcd(0, 0) = 0. gcd(a, 0) = a and gcd(0, b) = b because everything divides 0. If a and b are both even, gcd(a, b) = 2*gcd(a/2, b/2) because 2 is a common divisor. Multiplication with 2

Continue Reading “Steins GCD Algorithm” →

Almost Prime number

Asif NaeemSeptember 24, 2016

Problem:A no is said to be k-Almost Prime Number if it has exactly k prime factors (not necessary distinct). Your task is to complete the functionprintKAlmostPrimes which takes two argument k and N and prints the first N numbers that are k prime. Problem Link

/*You are required to complete this function*/

bool isPrime(int n)
{
    bool arr[20];
    int i;
    arr[2]=1;arr[3]=1;arr[5]=1;arr[7]=1;
    arr[11]=1;arr[13]=1;arr[17]=1;arr[19]=1;
    
    if(n<20)
    {
        if(arr[n]==1)
        return 1;
        else if(arr[n]==0)
        return 0;
    }
    if(n%2==0)
    return 0;
    for(i=3;i<=sqrt(n);i+=2)
    {
        if(n%i==0)
        return 0;
    }
    return 1;
}
int CountOfPrime(int n)
{
    int cnt=0,i;
    while(n%2==0)
    {
        cnt+=1;
        n/=2;
    }
    for(i=3;i<=sqrt(n);i+=2)
    {
        if(isPrime(i)==0)
        continue;
        while(n%i==0)
        {
            cnt+=1;
            n/=i;
        }
    }
    if(n>2)
    cnt+=1;
    return cnt;
}
void printKAlmostPrimes(int k, int n)
{
//Your code here
    int i,num;
    for(i=1,num=2;i<=n;num++)
    {
        if(CountOfPrime(num)==k)
        {
            printf("%d ",num);
            i++;
        }
    }
}

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/*You are required to complete this function*/

bool isPrime(int n)

{

bool arr[20];

int i;

arr[2]=1;arr[3]=1;arr[5]=1;arr[7]=1;

arr[11]=1;arr[13]=1;arr[17]=1;arr[19]=1;

if(n<20)

{

if(arr[n]==1)

return 1;

else if(arr[n]==0)

return 0;

}

if(n%2==0)

return 0;

for(i=3;i<=sqrt(n);i+=2)

{

if(n%i==0)

return 0;

}

return 1;

}

int CountOfPrime(int n)

{

int cnt=0,i;

while(n%2==0)

{

cnt+=1;

n/=2;

}

for(i=3;i<=sqrt(n);i+=2)

{

if(isPrime(i)==0)

continue;

while(n%i==0)

{

cnt+=1;

n/=i;

}

if(n>2)

cnt+=1;

return cnt;

}

void printKAlmostPrimes(int k, int n)

{

//Your code here

int i,num;

for(i=1,num=2;i<=n;num++)

{

if(CountOfPrime(num)==k)

{

printf("%d ",num);

i++;

}

10170

Asif NaeemApril 16, 2016

#include <iostream>
#include <cstdio>

#define ul unsigned long long//double//long long

using namespace std;

int main()
{
	ul s,d,a,b;
	ul subtract;
	ul lw,hi,mid,res;
	while(scanf("%llu%llu",&s,&d)!=EOF)
	{
		if(d<=s)
		{
			printf("%llu\n",s);
			continue;
		}
		subtract=(s*(s-1))/2;
		lw=s;
		hi=s+d/s;
		res=-1;
		while(hi-lw>1)
		{
			mid=(lw+hi)/2;
			if(((mid*(mid+1))/2-subtract) <d)
			lw=mid+1;
			else if(((mid*(mid+1))/2-subtract)>d)
			hi=mid;
			else if(((mid*(mid+1))/2-subtract)==d)
			{
				res=mid;
				break;
			}
		}
		if(res!=-1)
		printf("%llu\n",res);
		else
		{
			//cout<<lw<<" "<<hi<<"\n";
			a=(hi*(hi+1))/2-subtract;
			b=(lw*(lw+1))/2-subtract;
			if(a>=d && b<d)
			printf("%llu\n",hi);
			else
			printf("%llu\n",lw);
		}
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#define ul unsigned long long//double//long long

using namespace std;

int main()

{

ul s,d,a,b;

ul subtract;

ul lw,hi,mid,res;

while(scanf("%llu%llu",&s,&d)!=EOF)

{

if(d<=s)

{

printf("%llu\n",s);

continue;

}

subtract=(s*(s-1))/2;

lw=s;

hi=s+d/s;

res=-1;

while(hi-lw>1)

{

mid=(lw+hi)/2;

if(((mid*(mid+1))/2-subtract) <d)

lw=mid+1;

else if(((mid*(mid+1))/2-subtract)>d)

hi=mid;

else if(((mid*(mid+1))/2-subtract)==d)

{

res=mid;

break;

}

if(res!=-1)

printf("%llu\n",res);

else

{

//cout<<lw<<" "<<hi<<"\n";

a=(hi*(hi+1))/2-subtract;

b=(lw*(lw+1))/2-subtract;

if(a>=d && b<d)

printf("%llu\n",hi);

else

printf("%llu\n",lw);

}

return 0;

}

974

Asif NaeemApril 15, 2016

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>

#define ll long long

using namespace std;

ll arr[40001];
vector<int> vb;

ll _pow(int pwr)
{
	if(pwr==0)
	return 1;
	else
	return 10*_pow(pwr-1);
}

bool Isfunc(ll a,int b)
{
	int tmp;
	tmp=a;
	int pwr=0;
	int num=0;
	while(tmp)
	{
		num=num + (tmp%10)*_pow(pwr);
		tmp/=10;
		if(tmp+num==b && tmp!=0 && num!=0)
		return 1;
		pwr+=1;
	}
	return 0;
}

int main()
{
	int lw,hi;
	int t;
	int kase=1;
	vb.push_back(9);
	vb.push_back(45);
	vb.push_back(55);
	for(int i=91;i<=40000;i++)
	{
		if(Isfunc(i*i,i))
		vb.push_back(i);
	}
	//cout<<vb.size()<<"\n";
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&lw,&hi);
		printf("case #%d\n",kase++);
		bool found=0;
		for(int i=0;i<19;i++)
		{
			if(vb[i]>=lw && vb[i]<=hi)
			found=1,
			printf("%d\n",vb[i]);
		}
		if(found==0)
		printf("no kaprekar numbers\n");
		if(t)
		printf("\n");
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <vector>

#define ll long long

using namespace std;

ll arr[40001];

vector<int> vb;

ll _pow(int pwr)

{

if(pwr==0)

return 1;

else

return 10*_pow(pwr-1);

}

bool Isfunc(ll a,int b)

{

int tmp;

tmp=a;

int pwr=0;

int num=0;

while(tmp)

{

num=num + (tmp%10)*_pow(pwr);

tmp/=10;

if(tmp+num==b && tmp!=0 && num!=0)

return 1;

pwr+=1;

}

return 0;

}

int main()

{

int lw,hi;

int t;

int kase=1;

vb.push_back(9);

vb.push_back(45);

vb.push_back(55);

for(int i=91;i<=40000;i++)

{

if(Isfunc(i*i,i))

vb.push_back(i);

}

//cout<<vb.size()<<"\n";

scanf("%d",&t);

while(t--)

{

scanf("%d%d",&lw,&hi);

printf("case #%d\n",kase++);

bool found=0;

for(int i=0;i<19;i++)

{

if(vb[i]>=lw && vb[i]<=hi)

found=1,

printf("%d\n",vb[i]);

}

if(found==0)

printf("no kaprekar numbers\n");

if(t)

printf("\n");

}

return 0;

}

594

Asif NaeemApril 14, 2016

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>

using namespace std;

int main()
{
	int n,res;
	int positionInByte,byteNo;
	while(scanf("%d",&n)!=EOF)
	{
		res=0;
		for(int i=0;i<32;i++)
		{
			if((1<<i)&n)
			{
				byteNo=i/8;
				positionInByte=i%8;
				res |= (1<<((3-byteNo)*8+positionInByte));
			}
		}
		printf("%d converts to %d\n",n,res);
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

using namespace std;

int main()

{

int n,res;

int positionInByte,byteNo;

while(scanf("%d",&n)!=EOF)

{

res=0;

for(int i=0;i<32;i++)

{

if((1<<i)&n)

{

byteNo=i/8;

positionInByte=i%8;

res |= (1<<((3-byteNo)*8+positionInByte));

}

printf("%d converts to %d\n",n,res);

}

return 0;

}

471

Asif NaeemApril 10, 2016

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>

#define ll long long
#define LIMIT 9876543210

using namespace std;


vector<int> vb;

bool isRepeated(ll num)
{
	if(num>=0 && num<=10)
	return 0;
	bool digits[10];
	memset(digits,0,sizeof(digits));
	while(num)
	{
		if(digits[num%10])
		return 1;
		digits[num%10]=1;
		num/=10;
	}
	return 0;
}
int main()
{
	int t;
	ll n;
	ll num;
	for(int i=1;i<1000005;i++)
	{
		if(isRepeated(i)==0)
		vb.push_back(i);
	}
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld",&n);
		for(int i=0;i<vb.size();i++)
		{
			num=vb[i]*n;
			if(num>LIMIT)
			{
				break;
			}
			if(isRepeated(num)==0)
			{
				printf("%lld / %d = %lld\n",num,vb[i],n);
			}
		}
		if(t)
		printf("\n");
	}
}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <vector>

#define ll long long

#define LIMIT 9876543210

using namespace std;

vector<int> vb;

bool isRepeated(ll num)

{

if(num>=0 && num<=10)

return 0;

bool digits[10];

memset(digits,0,sizeof(digits));

while(num)

{

if(digits[num%10])

return 1;

digits[num%10]=1;

num/=10;

}

return 0;

}

int main()

{

int t;

ll n;

ll num;

for(int i=1;i<1000005;i++)

{

if(isRepeated(i)==0)

vb.push_back(i);

}

scanf("%d",&t);

while(t--)

{

scanf("%lld",&n);

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

{

num=vb[i]*n;

if(num>LIMIT)

{

break;

}

if(isRepeated(num)==0)

{

printf("%lld / %d = %lld\n",num,vb[i],n);

}

if(t)

printf("\n");

}

107

Asif NaeemApril 5, 2016

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>

#define ll long long
#define MOD 1000000007
using namespace std;

int _pow(int n,int p)
{
	if(p==0)
		return 1;
	else
		return n*_pow(n,p-1);
}

int isAnyRoot(int a,int b)
{
	if(a%b!=0)
		return -1;
	int cnt=0;
	while(a>=b)
	{
		cnt+=1;
		a=a/b;
		if(a>=b && a%b!=0)
			return -1;
	}
	return cnt;
}
int main()
{
	//cout<<_pow(6,2)<<"\n";
	int H,W,m,n,non_working_cat;
	int c1,c2;
	int total_non_working_cat , height_of_the_cat_stack , height_of_each_internal_cat;
	while(scanf("%d%d",&H,&W),H!=0,W!=0)
	{
		if(H==1)
		{
			printf("%d %d\n",0,H);
			continue;
		}
		//that means there is one working cat ; so there can be two possible scenarios
		//1. if the height of the initial cat is nth power of 2,
		//i.e. just next(inside of which hat working cat is staying) non working cat
		//whose height should be 2 so that 2/N+1==1;
		// In this acse total non working cat will be n (2^n=initial height), and height of the cat stack = 2+4+...2^n
		//2. if the height of the initial cat is not nth power of 2; then 1=(Height)/N+1, so N+1=Height,
		//means total cat= Height of the initial cat
		//only one is non working rest are working cat
		//and total height should be (N*1+initial height) or (N*1+1) if they are considered in a same ground
		if(W==1)
		{
			m = isAnyRoot(H,2);
			if(m!=-1)
			{
				printf("%d %d\n",m,_pow(2,m+1)-1);
			}
			else
			{
				printf("%d %d\n",1,2*H-1);
			}
			continue;
		}
		/*
		m=0;
		while(++m)
		{
			n = pow(W*1.0,1.0/m);
			if(H<=pow(n+1, m))
			break;
		}*/
		n=1;
		while(1)
		{
			n+=1;
			c1 = isAnyRoot(W,n);
			c2 = isAnyRoot(H,n+1);
			if(c1==c2 && c1!=-1)
				break;
		}
		m=c1;
		total_non_working_cat = 0;
		height_of_the_cat_stack = 0;
		for(int i=0;i<m;i++)
		{
			non_working_cat = _pow(n,i);
			total_non_working_cat += non_working_cat;
			height_of_each_internal_cat = (H/_pow(n+1,i));
			height_of_the_cat_stack += (non_working_cat*height_of_each_internal_cat);
		}
		height_of_the_cat_stack += W;//as working cats are of height 1.
		printf("%d %d\n",total_non_working_cat,height_of_the_cat_stack);
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#define ll long long

#define MOD 1000000007

using namespace std;

int _pow(int n,int p)

{

if(p==0)

return 1;

else

return n*_pow(n,p-1);

}

int isAnyRoot(int a,int b)

{

if(a%b!=0)

return -1;

int cnt=0;

while(a>=b)

{

cnt+=1;

a=a/b;

if(a>=b && a%b!=0)

return -1;

}

return cnt;

}

int main()

{

//cout<<_pow(6,2)<<"\n";

int H,W,m,n,non_working_cat;

int c1,c2;

int total_non_working_cat , height_of_the_cat_stack , height_of_each_internal_cat;

while(scanf("%d%d",&H,&W),H!=0,W!=0)

{

if(H==1)

{

printf("%d %d\n",0,H);

continue;

}

//that means there is one working cat ; so there can be two possible scenarios

//1. if the height of the initial cat is nth power of 2,

//i.e. just next(inside of which hat working cat is staying) non working cat

//whose height should be 2 so that 2/N+1==1;

// In this acse total non working cat will be n (2^n=initial height), and height of the cat stack = 2+4+...2^n

//2. if the height of the initial cat is not nth power of 2; then 1=(Height)/N+1, so N+1=Height,

//means total cat= Height of the initial cat

//only one is non working rest are working cat

//and total height should be (N*1+initial height) or (N*1+1) if they are considered in a same ground

if(W==1)

{

m = isAnyRoot(H,2);

if(m!=-1)

{

printf("%d %d\n",m,_pow(2,m+1)-1);

}

else

{

printf("%d %d\n",1,2*H-1);

}

continue;

}

m=0;

while(++m)

{

n = pow(W*1.0,1.0/m);

if(H<=pow(n+1, m))

break;

}*/

n=1;

while(1)

{

n+=1;

c1 = isAnyRoot(W,n);

c2 = isAnyRoot(H,n+1);

if(c1==c2 && c1!=-1)

break;

}

m=c1;

total_non_working_cat = 0;

height_of_the_cat_stack = 0;

for(int i=0;i<m;i++)

{

non_working_cat = _pow(n,i);

total_non_working_cat += non_working_cat;

height_of_each_internal_cat = (H/_pow(n+1,i));

height_of_the_cat_stack += (non_working_cat*height_of_each_internal_cat);

}

height_of_the_cat_stack += W;//as working cats are of height 1.

printf("%d %d\n",total_non_working_cat,height_of_the_cat_stack);

}

return 0;

}

11827

Asif NaeemMarch 19, 2016

#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

vector<int> vb;

int gcd(int a,int b)
{
	if(a%b==0)
	return b;
	else
	return gcd(b,a%b);
}
int main()
{
	int n,a;
	int x,y;
	string str;
	cin>>n;
	getchar();
	stringstream iss;
	int mx;
	while(n--)
	{
		//cin>>x>>y;
		//cout<<gcd(x,y)<<"\n";
		getline(cin,str);
		iss<<str;
		mx=0;
		vb.clear();
		while(iss>>a)
		{
			//cout<<"sdd\n";
			vb.push_back(a);
		}
		iss.clear();
		for(int i=0;i<vb.size()-1;i++)
		{
			for(int j=i+1;j<vb.size();j++)
			{
				a = gcd(vb[i],vb[j]);
				if(a>mx)
				mx = a;
			}
		}
		printf("%d\n",mx);
	}
	return 0;
}

#include <iostream>

#include <sstream>

#include <cstdio>

#include <cstring>

#include <vector>

using namespace std;

vector<int> vb;

int gcd(int a,int b)

{

if(a%b==0)

return b;

else

return gcd(b,a%b);

}

int main()

{

int n,a;

int x,y;

string str;

cin>>n;

getchar();

stringstream iss;

int mx;

while(n--)

{

//cin>>x>>y;

//cout<<gcd(x,y)<<"\n";

getline(cin,str);

iss<<str;

mx=0;

vb.clear();

while(iss>>a)

{

//cout<<"sdd\n";

vb.push_back(a);

}

iss.clear();

for(int i=0;i<vb.size()-1;i++)

{

for(int j=i+1;j<vb.size();j++)

{

a = gcd(vb[i],vb[j]);

if(a>mx)

mx = a;

}

printf("%d\n",mx);

}

return 0;

}

10223

Asif NaeemMarch 14, 2016

#include <iostream>
#include <cstdio>

#define ll long long

using namespace std;

ll catalan[25];

ll findcatalan(int n)
{
	ll res = 1;
	int lim;
	for(int i=n+2;i<=2*n;i++)
	{
		if(i%2==0)
		res*=2;
		else
		res*=i;
	}
	if((n+2)%2==0)
	lim = (n+2)/2;
	else
	lim = (n+3)/2;
	for(int i=2;i<lim;i++)
	{
		res/=i;
	}
	return res;
}

int binSearch(ll num)
{
    int lw=0,hi=17;
    int mid;
    /*while(lw<hi)
    {
        mid=lw+(hi-lw)/2;
        if(catalan[mid]<num)
        lw = mid+1;
        else if(catalan[mid]==num)
        return mid;
        else if(catalan[mid]>num)
        hi = mid;
    }*/
    for(int i=1;i<20;i++)
    {
            if(num==catalan[i])
            return i;
    }
    return 0;
}
int main()
{
	catalan[0]=1;
	catalan[1]=1;
	catalan[2]=2;
	ll num;
	int idx;
	for(int i=3;i<20;i++)
	{
		catalan[i] = findcatalan(i);
	}
	while(cin>>num)
	{
		idx = binSearch(num);
		cout<<idx<<"\n";
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#define ll long long

using namespace std;

ll catalan[25];

ll findcatalan(int n)

{

ll res = 1;

int lim;

for(int i=n+2;i<=2*n;i++)

{

if(i%2==0)

res*=2;

else

res*=i;

}

if((n+2)%2==0)

lim = (n+2)/2;

else

lim = (n+3)/2;

for(int i=2;i<lim;i++)

{

res/=i;

}

return res;

}

int binSearch(ll num)

{

int lw=0,hi=17;

int mid;

/*while(lw<hi)

{

mid=lw+(hi-lw)/2;

if(catalan[mid]<num)

lw = mid+1;

else if(catalan[mid]==num)

return mid;

else if(catalan[mid]>num)

hi = mid;

}*/

for(int i=1;i<20;i++)

{

if(num==catalan[i])

return i;

}

return 0;

}

int main()

{

catalan[0]=1;

catalan[1]=1;

catalan[2]=2;

ll num;

int idx;

for(int i=3;i<20;i++)

{

catalan[i] = findcatalan(i);

}

while(cin>>num)

{

idx = binSearch(num);

cout<<idx<<"\n";

}

return 0;

}

Category: Math

Karatsuba’s Fast Multiplication Algorithm

Steins GCD Algorithm

Almost Prime number

10170

974

594

471

107

11827

10223

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