July 2015 – Page 4 – Live with passion

10048

Asif NaeemJuly 8, 2015

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

#define INF 1000000
using namespace std;

int gr[105][105];

int _mn(int a,int b)
{
	if(a>b)
	return b;
	return a;
}

int _mx(int a,int b)
{
	if(a>b)
	return a;
	return b;
}
int main()
{
	bool f=0;
	int kase=1;
	int C,S,Q;
	int a,b,d;

	while(cin>>C)
	{
		cin>>S>>Q;

		if(C+S+Q==0)
		break;

		for(int i=1;i<105;i++)
		{
			for(int j=1;j<105;j++)
			{
				if(i==j)
				gr[i][j]=0;
				else
				gr[i][j]=INF;
			}
		}
		for(int i=1;i<=S;i++)
		{
			cin>>a>>b>>d;
			gr[a][b]=d;
			gr[b][a]=d;
		}

		for(int i=1;i<=C;i++)
		{
			for(int j=1;j<=C;j++)
			{
				for(int k=1;k<=C;k++)
				{
					gr[j][k]=_mn(gr[j][k],_mx(gr[j][i],gr[i][k]));
				}
			}
		}

		if(f)
		cout<<"\n";
		f=1;
		cout<<"Case #"<<kase++<<"\n";
		for(int i=0;i<Q;i++)
		{
			cin>>a>>b;
			if(gr[a][b]==INF)
			cout<<"no path\n";
			else
			cout<<gr[a][b]<<"\n";
		}
	}
	return 0;
}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#define INF 1000000

using namespace std;

int gr[105][105];

int _mn(int a,int b)

{

if(a>b)

return b;

return a;

}

int _mx(int a,int b)

{

if(a>b)

return a;

return b;

}

int main()

{

bool f=0;

int kase=1;

int C,S,Q;

int a,b,d;

while(cin>>C)

{

cin>>S>>Q;

if(C+S+Q==0)

break;

for(int i=1;i<105;i++)

{

for(int j=1;j<105;j++)

{

if(i==j)

gr[i][j]=0;

else

gr[i][j]=INF;

}

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

{

cin>>a>>b>>d;

gr[a][b]=d;

gr[b][a]=d;

}

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

{

for(int j=1;j<=C;j++)

{

for(int k=1;k<=C;k++)

{

gr[j][k]=_mn(gr[j][k],_mx(gr[j][i],gr[i][k]));

}

if(f)

cout<<"\n";

f=1;

cout<<"Case #"<<kase++<<"\n";

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

{

cin>>a>>b;

if(gr[a][b]==INF)

cout<<"no path\n";

else

cout<<gr[a][b]<<"\n";

}

return 0;

}

Golbach’s Conjecture: Any even number can be expressed as a summation of two prime numbers. This idea is used to solve the problem. if the number is odd then .ie 49 thn 2,3 and goldbach conjechture of (49-5) can be expressed as sum of 2 primes. if the number is even then 2,2 and goldbach

Continue Reading “10168” →

729

Asif NaeemJuly 7, 2015

Using Back Track Algorithm:

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

using namespace std;
int N,H;
int backtrack(int idx,int cnt,int a[20])
{
	if(cnt==N-H)
	{
		for(int i=1;i<=N;i++)
			cout<<a[i];
		cout<<"\n";
		return 0;
	}
	for(int i=idx;i<=N;i++)
	{
		a[i]=0;
		backtrack(i+1,cnt+1,a);
		a[i]=1;
	}
	return 0;
}

int main()
{
	int t;
	string s;
	cin>>t;
	getchar();
	getline(cin,s);
	int str[20];
	while(t--)
	{
		cin>>N;
		getchar();
		cin>>H;
		getchar();
		getline(cin,s);
		for(int i=0;i<20;i++)
			str[i]=1;
		backtrack(1,0,str);
		if(t)
		cout<<"\n";

	}
	return 0;
}

#include <iostream>

#include <cstring>

#include <cstdio>

#include <string>

#include <algorithm>

using namespace std;

int N,H;

int backtrack(int idx,int cnt,int a[20])

{

if(cnt==N-H)

{

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

cout<<a[i];

cout<<"\n";

return 0;

}

for(int i=idx;i<=N;i++)

{

a[i]=0;

backtrack(i+1,cnt+1,a);

a[i]=1;

}

return 0;

}

int main()

{

int t;

string s;

cin>>t;

getchar();

getline(cin,s);

int str[20];

while(t--)

{

cin>>N;

getchar();

cin>>H;

getchar();

getline(cin,s);

for(int i=0;i<20;i++)

str[i]=1;

backtrack(1,0,str);

if(t)

cout<<"\n";

}

return 0;

}

Using Next_permutation() method in C++

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

using namespace std;
int N,H;

int main()
{
	int t;
	string s;
	cin>>t;
	getchar();
	getline(cin,s);
	int str[20];
	while(t--)
	{
		cin>>N;
		getchar();
		cin>>H;
		getchar();
		getline(cin,s);
		for(int i=0;i<N-H;i++)
			str[i]=0;
		for(int i=N-H;i<N;i++)
			str[i]=1;
		do{
			for(int i=0;i<N;i++)
				cout<<str[i];
			cout<<"\n";
		}
		while(next_permutation(str,str+N));
		if(t)
		cout<<"\n";

	}
	return 0;
}

#include <iostream>

#include <cstring>

#include <cstdio>

#include <string>

#include <algorithm>

using namespace std;

int N,H;

int main()

{

int t;

string s;

cin>>t;

getchar();

getline(cin,s);

int str[20];

while(t--)

{

cin>>N;

getchar();

cin>>H;

getchar();

getline(cin,s);

for(int i=0;i<N-H;i++)

str[i]=0;

for(int i=N-H;i<N;i++)

str[i]=1;

do{

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

cout<<str[i];

cout<<"\n";

}

while(next_permutation(str,str+N));

if(t)

cout<<"\n";

}

return 0;

}

138

Asif NaeemJuly 7, 2015

Theory: Perfect square problem.Problem is to determine a pair of numbers where the small (lower_num) number divides the numbers from 1 up to higher_num into two groups in a way that summation of 1 to (lower_num-1) is equal to summation of the numbers from (lower_num+1) to higher_num. Expanation: first pair 6 8 summation of 1

Continue Reading “138” →

10193

Asif NaeemJuly 7, 2015

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

using namespace std;

int gcd(int a,int b)
{
	if(a%b==0)
	return b;
	else
	return gcd(b,a%b);
}
int main()
{
	int t,kase=1,num1,num2;
	string s1,s2;
//	cin>>num1>>num2;
//	cout<<gcd(num1,num2)<<"\n";
	cin>>t;

	while(t--)
	{
		cin>>s1;
		cin>>s2;
		num1=0;
		num2=0;
		int prev=1;
		for(int i=s1.size()-1;i>=0;i--)
		{
			num1=num1+(s1[i]-'0')*prev;
			prev*=2;
		}
		prev=1;
		for(int i=s2.size()-1;i>=0;i--)
		{
			num2=num2+(s2[i]-'0')*prev;
			prev*=2;
		}

		//cout<<gcd(num1,num2)<<"\n";
		if(gcd(num1,num2)==1)
		cout<<"Pair #"<<kase++<<": Love is not all you need!\n";
		else
		cout<<"Pair #"<<kase++<<": All you need is love!\n";
	}
	return 0;
}

#include <iostream>

#include <cstring>

#include <cstdio>

#include <string>

#include <algorithm>

using namespace std;

int gcd(int a,int b)

{

if(a%b==0)

return b;

else

return gcd(b,a%b);

}

int main()

{

int t,kase=1,num1,num2;

string s1,s2;

// cin>>num1>>num2;

// cout<<gcd(num1,num2)<<"\n";

cin>>t;

while(t--)

{

cin>>s1;

cin>>s2;

num1=0;

num2=0;

int prev=1;

for(int i=s1.size()-1;i>=0;i--)

{

num1=num1+(s1[i]-'0')*prev;

prev*=2;

}

prev=1;

for(int i=s2.size()-1;i>=0;i--)

{

num2=num2+(s2[i]-'0')*prev;

prev*=2;

}

//cout<<gcd(num1,num2)<<"\n";

if(gcd(num1,num2)==1)

cout<<"Pair #"<<kase++<<": Love is not all you need!\n";

else

cout<<"Pair #"<<kase++<<": All you need is love!\n";

}

return 0;

}

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