### hi hamedv

Posted:

**Wed Jul 25, 2007 7:20 am**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.

Thank you for your reply.

Page **5** of **8**

Posted: **Wed Jul 25, 2007 7:20 am**

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.

Thank you for your reply.

Posted: **Wed Jul 25, 2007 7:39 am**

mr ishtiaqq,

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

if you replaced:

it will work good as you want.

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

if you replaced:

Code: Select all

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

Posted: **Wed Jul 25, 2007 11:24 am**

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

Code: Select all

```
#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;
}
```

Posted: **Wed Jul 25, 2007 2:13 pm**

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

Posted: **Wed Jul 25, 2007 4:46 pm**

Jan bhia was written
Let there are 10 data in the array as follows
here
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

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

Code: Select all

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

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

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

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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
```

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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]
```

Code: Select all

```
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;
```

Posted: **Sat May 17, 2008 8:53 am**

Someone can plesae tell me why I'm getting TLE ?

I read previous posts.

Here is my code :
Thank's in advance.

I read previous posts.

Here is my code :

Code: Select all

```
#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;
}
```

Posted: **Sun May 18, 2008 10:39 am**

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

Something like this -

Something like this -

Code: Select all

```
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;
}
}
}
}
```

Posted: **Sun May 18, 2008 12:56 pm**

To emotional Blind,

I did as you said but got TLE again with running time 3.00 sec like past.

I did it as :
Thanks for reply.

I did as you said but got TLE again with running time 3.00 sec like past.

I did it as :

Code: Select all

```
Removed after AC
```

Posted: **Mon May 19, 2008 3:26 am**

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

Posted: **Mon May 19, 2008 3:38 pm**

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.

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.

Posted: **Tue May 20, 2008 4:21 am**

You are welcome.mukit wrote:Thank's a lot emotional blind. I got AC with 0.56 sec.

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

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

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.

Posted: **Wed Jul 02, 2008 1:24 pm**

Ishtiaq Ahmed wrote:

Edit!!!!!!Jan bhia was written

Posted: **Wed Jul 02, 2008 1:30 pm**

emotional blind wrote:

Would someone please tell about the use of memset()?

Why cin and cout are costly?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.

Would someone please tell about the use of memset()?

- The header file.

How to use it.

Posted: **Mon Jul 21, 2008 12:23 pm**

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...

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...

Code: Select all

```
#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");
}
```

Posted: **Mon Jul 28, 2008 11:44 am**

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

Code: Select all

`removed`