10533 - Digit Primes

[ishtiaq ahmed]

i think its not needed to close the bracket as you have said. It works the same as you think.

Thank you for your reply.

[newton]

mr ishtiaqq,

you needn't count upto i<SIZE for the outer loop.
if you replaced:

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the condition i<SIZE
into:
i<= sqrt(SIZE)

it will work good as you want.

[ishtiaq ahmed]

I have updated my code as follows. Firstly it faced TLE and then as follows. Please try to help me.

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#include<stdio.h>
#include<math.h>

#define SIZE 1000001

long  a,b,data[SIZE]={0},add[SIZE]={0};


void s (void);
long sum_prime(void);


int main()
{
	long k,i,result,cas,sum=0,z,dum,m;
	s();
	sum=0;
	for(i=0;i<SIZE;i++)
	{
		if(!data[i])
		{
			z=i;dum=0;
			while(z)
			{
				m=z%10;
				dum +=m;
				z /=10;
			}
			if(!data[dum])
			{
				sum++;
				add[i]=sum;
			}
		}
		else
			add[i]=sum;
	}
	scanf("%ld",&cas);
	for(k=1;k<=cas;k++)
	{
		scanf("%ld %ld",&a, &b);
		result=sum_prime();
		printf("%ld\n",result);
	}
	return 0;
}



void s(void)
{
	long i,j,sqrtNum;
	data[0]=1;
	data[1]=1;
	for(i=4;i<SIZE;i+=2)
		data[i]=1;
	sqrtNum = (long)sqrt(SIZE);
	for(i=3;i<sqrtNum;i+=2)
		for(j=i;i*j<SIZE;j++)
			if(!data[i*j])
				data[i*j]=1;
}



long sum_prime(void)
{
	long XX,z,dum,m;
	if(a!=b)
		XX=add[b]-add[a];
	else if(a==b)
	{
		if(!data[a])
		{
			dum=0;z=a;
			while(z)
			{
				m=z%10;
				dum +=m;
				z /=10;
			}
			if(!data[dum])
				XX=1;
		}
		else
			XX=0;
	}
	return XX;
}

[Jan]

What if you use add - add[a-1] ?

[ishtiaq ahmed]

Jan bhia was written

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What if you use add[b] - add[a-1] ?

Let there are 10 data in the array as follows

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            1  2  3  4  5  6  7  8  9  10
            |  |  |  |  |  |  |  |  |  |
            N  P  P  N  P  N  P  N  N  N

here

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     N-> Not prime
     P-> Prime & have prime digit

Now i consider that if data is prime and the prime data's summation of didits are prime then i add them just like these

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            1  2  3  4  5  6  7  8  9  10
            |  |  |  |  |  |  |  |  |  |
            N  P  P  N  P  N  P  N  N  N
     add[]->0  1  2  2  3  3  4  4  4  4

Now when i am asked to find out the total summation that is asked by this problem tryed to solve like this:-

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Question: Find the total answer from 5th[a] element to 10th[b] element?
answer:  add[b-1] - add[a-1] that is 
         add[9] - add[4]

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Question: Find the total answer from 4th[a] element to 4th[b] element?

answer: First i have checked if the 4th element is prime and have prime digit then i printed 1 otherwise 0;  

[mukit]

Someone can plesae tell me why I'm getting TLE ?
I read previous posts.
Here is my code :

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#include<iostream>
#include<cstdio>

using namespace std;

#define SIZE 1000002

long  a,b,i,j;
char data[SIZE]={0};
int dp[SIZE];
int nop[SIZE];
void sieve(void)
{
   int i, j, k;

   data[0] = 1;
   data[1] = 1;

   for (i=4; i<SIZE; i+=2)
   {
      data[i] = 1;
   }

   for (i=3; i<SIZE; i+=2)
   {
      if (!data[i])
      {
         k = SIZE / i;
      
         for (j=i; j<=k; j+=2)
         {
            data[i * j] = 1;
         }
      }

   
   }
}
void is_dp()
{
	long x,z,ct=0,du,m=0;
	for(i=0;i<SIZE;i++)
	{
		if(!data[i])
		{
			z=x=i;
			du=0;
			while(z)
			{
				m=z%10;
				du+=m;
				z/=10;
			}
			if(!data[du])
			{
				dp[i]=1;
			}
		}
	}
}
void sum_prime()
{
	int sum=0,dum,m=0,z;
	for(i=0;i<=SIZE;i++)
	{
		if(dp[i])
		{
			sum+=1;
		}
		nop[i]=sum;
	}
}
int main()
{
	long test;
	sieve();
	is_dp();
	sum_prime();
	cin >> test;
	for(long i=0;i<test;i++) 
	{
		cin>>a>>b;
		cout<<nop[b]-nop[a-1]<<endl;
	}
	return 0;
}

Thank's in advance.

[emotional blind]

probably your is_dp function is costly. try to re-implement this is_dp using something real dp
Something like this -

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	int ds[SIZE];
	int dsum(int n){
		int &ret = ds[n];
		if(-1!=ret)return ret;

		if(n<10)return ret=n;

		return ret = n%10 + dsum(n/10);
	}

	void is_dp(){
		long x,z,ct=0,du,m=0;
       memset(ds,-1,sizeof(ds));

		for(i=0;i<SIZE;i++)
       {
          if(!data[i])
          {
             du = dsum(i);
			 if(!data[du])
             {
                dp[i]=1;
             }
          }
       }	
	}

[mukit]

To emotional Blind,
I did as you said but got TLE again with running time 3.00 sec like past.
I did it as :

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Removed after AC

Thanks for reply.

[emotional blind]

now, try again with replacing all cin-cout with scanf-printf.

[mukit]

Thank's a lot emotional blind. I got AC with 0.56 sec.
But something was ambiguous to me. It didn't find the memset() function in ANSI C with header #include<stdio.h> and
only #include<cstdio> header in C++. I had to use #include<iostream> with namespace.
Another thing why it took such a huge time(3.00) with cin-cout getting TLE?

After all thank you very much.

[emotional blind]

Thank's a lot emotional blind. I got AC with 0.56 sec.

You are welcome.

It didn't find the memset() function in ANSI C with header #include<stdio.h>

#include <string.h>

Another thing why it took such a huge time(3.00) with cin-cout getting TLE?

cin and cout is lot more costly than scanf and printf, this is noticeable for the problems which needs large input and output data to handle.
I think it takes more than 3.00 seconds. though 3.00 is the time limit, after 3.00 seconds the judge system terminates your program.

[lnr]

Ishtiaq Ahmed wrote:

Jan bhia was written

Edit!!!!!!

[lnr]

emotional blind wrote:

cin and cout is lot more costly than scanf and printf, this is noticeable for the problems which needs large input and output data to handle.
I think it takes more than 3.00 seconds. though 3.00 is the time limit, after 3.00 seconds the judge system terminates your program.

Why cin and cout are costly?
Would someone please tell about the use of memset()?

• The header file.
  How to use it.

[shekhar]

PLZZ Help...I am Getting TLE in this problem
I used sieve for prime & dprime.Current sieve takes around 0.2-0.3 seconds for calculation.Can anyone help me out...how optimization can be done??
Here is my code...

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#include<iostream>
#include<cmath>
using namespace std;
bool prime[1000001];
bool dprime[1000001];
void seive()
{
     int m=1000;
     memset(prime,true,sizeof(prime));
     prime[0]=false;
     prime[1]=false;
     for(int i=2;i<=m;i++)
     if(prime[i]) 
     for(int k=i*i;k<=1000000;k+=i)
                  prime[k]=false;
                  
     memset(dprime,false,sizeof(dprime));
     dprime[0]=false;
     dprime[1]=false;
     for(int j=2;j<=1000000;j++)
     {
             int sum=0,num=j;
             while(num>=1)
             {
                          sum+=num%10;
                          num=num/10;
             }
             if(prime[sum])
             dprime[j]=true;
     }        
}
int main()
{
    int test,a,b;
    seive();
    scanf("%d",&test);
    while(test--)
    {
              scanf("%d %d",&a,&b);
              int n,count=0;
              for(n=a;n<=b;n++)
              {
                               if(prime[n] && dprime[n])
                               count++;                              
              }
              printf("%d\n",count);
    }
    system("pause");
}

[Obaida]

Some one please help me to get Acc less TLE. I tried but failed so many times.

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removed