10948  The primary problem
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10948  The primary problem.
I think this problem is prety easy, but i am getting WA.
plaese help with some tricky input.
Thanks
plaese help with some tricky input.
Thanks
Junayeed
Nothing tricky, just some boundary cases:
Output:
Code: Select all
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
999980
999981
999982
999983
999984
999985
999986
999987
999988
999989
999990
999991
999992
999993
999994
999995
999996
999997
999998
999999
1000000
0
Code: Select all
4:
2+2
5:
2+3
6:
3+3
7:
2+5
8:
3+5
9:
2+7
10:
3+7
11:
NO WAY!
12:
5+7
13:
2+11
14:
3+11
15:
2+13
16:
3+13
17:
NO WAY!
18:
5+13
19:
2+17
20:
3+17
999980:
19+999961
999981:
2+999979
999982:
3+999979
999983:
NO WAY!
999984:
5+999979
999985:
2+999983
999986:
3+999983
999987:
NO WAY!
999988:
5+999983
999989:
NO WAY!
999990:
7+999983
999991:
NO WAY!
999992:
13+999979
999993:
NO WAY!
999994:
11+999983
999995:
NO WAY!
999996:
13+999983
999997:
NO WAY!
999998:
19+999979
999999:
NO WAY!
1000000:
17+999983
10948  The primary problem
someone pls help me out. im getting wa with this code, i've tried so many test cases, all the outputs r right. but still gettin wa. pls help.
#include<iostream.h>
#include<math.h>
#define P 1000002
bool prime[P];
int main()
{
int i,j;
prime[0]=prime[1]=1;
for(i=2; i<=sqrt(P1); i++)
{
if(prime==0)
{
for(j=i*i; j<=P1; j=j+i)
{
prime[j]=1;
}
}
}
while(cin>>i)
{
if(i==0) break;
for(j=2;j<=i/2;j++)
{
if( (prime[j]==0) && (prime[ij]==0) ) break;
}
if(j>(i/2)) cout<<"NO WAY!\n";
else
{
cout<<i<<":\n"<<j<<"+"<<(ij)<<"\n";
}
}
return 0;
}
#include<iostream.h>
#include<math.h>
#define P 1000002
bool prime[P];
int main()
{
int i,j;
prime[0]=prime[1]=1;
for(i=2; i<=sqrt(P1); i++)
{
if(prime==0)
{
for(j=i*i; j<=P1; j=j+i)
{
prime[j]=1;
}
}
}
while(cin>>i)
{
if(i==0) break;
for(j=2;j<=i/2;j++)
{
if( (prime[j]==0) && (prime[ij]==0) ) break;
}
if(j>(i/2)) cout<<"NO WAY!\n";
else
{
cout<<i<<":\n"<<j<<"+"<<(ij)<<"\n";
}
}
return 0;
}
Re: 10948 pls help
Hi,
your code didn't return N itself as a part of output, if N cannot be represented as the sum of two prime number.
Please check your output format carefully.
Best regards.
your code didn't return N itself as a part of output, if N cannot be represented as the sum of two prime number.
Please check your output format carefully.
Best regards.

 Learning poster
 Posts: 53
 Joined: Sat Jul 29, 2006 7:33 am
 Location: (CSE,DU), Dhaka,Bangladesh
10948(Primary Problem)TLE
Can anybody help me? I am facing TLE (10.035s). Here is my code
Please try to help me.
Code: Select all
cut after ac
Last edited by ishtiaq ahmed on Fri Jul 27, 2007 4:00 pm, edited 1 time in total.
No venture no gain
with best regards

ishtiaq ahmed
with best regards

ishtiaq ahmed
Code: Select all
if(!flag)
{
for(i=temp;i>=(number/2);i=2)
{
if(!data[i])
{
for(j=3;i+j<=number;j+=2) < ??? why do you need this loop ?
{
if(!data[j] && i+j==number)
{
flag=1;
printf("%ld:\n",number);
printf("%ld+%ld\n",j,i);
break;
}
Rio

 Learning poster
 Posts: 53
 Joined: Sat Jul 29, 2006 7:33 am
 Location: (CSE,DU), Dhaka,Bangladesh
post reply of previous submission of 10948
Thanks rio for replying me.
At first i have generated the prime number upto 1000000. When the input is 18 output will
Firstly i have checked any given number that is greater than 4 even or not. if it is even subtracted it with 1, otherwise 2.
so the number is now 17(temporarily).
then i checked if 2+17==18? if not then enter The first for loop. first for loop handled data from 17 to ...............>(18/2);and as well as the second one that is asked by rio handled data from 3 to(18/2).
Now let see..
So these for two for loops are incremented or decremented by 2.
I think you understand my algorithm. If there any efficient algorithm let me know and also try to find my errors in this code.
At first i have generated the prime number upto 1000000. When the input is 18 output will
Code: Select all
18 = 5 + 13;
so the number is now 17(temporarily).
then i checked if 2+17==18? if not then enter The first for loop. first for loop handled data from 17 to ...............>(18/2);and as well as the second one that is asked by rio handled data from 3 to(18/2).
Now let see..
Code: Select all
3+17 == 18 (no) (3+17=20 exceeds 18) then
skip checking 3+15== 18. Because 15 is not prime. then
3+13 == 18 (no) (3+13=16< 18)so
5+13 == 18. Then stop and print the result.
I think you understand my algorithm. If there any efficient algorithm let me know and also try to find my errors in this code.
Last edited by ishtiaq ahmed on Fri Jul 27, 2007 4:01 pm, edited 1 time in total.
No venture no gain
with best regards

ishtiaq ahmed
with best regards

ishtiaq ahmed
Yes, I could understand your algorithm, and it is efficient enough.
Few things to say.
1. Actually, you don't need the loop which I mentioned.
Look careful. ifblock is executed at most once, when j = number  i is a prime.
This waste loop is causing TLE.
2. Its not "NO WAY", but "NO WAY!"
3. 1 is not a prime.
Hope this helps.

Rio
Few things to say.
1. Actually, you don't need the loop which I mentioned.
Code: Select all
for(j=3;i+j<=number;j+=2)
{
if(!data[j] && i+j==number)
{
flag=1;
printf("%ld:\n",number);
printf("%ld+%ld\n",j,i);
break;
}
}
This waste loop is causing TLE.
2. Its not "NO WAY", but "NO WAY!"
3. 1 is not a prime.
Hope this helps.

Rio

 Learning poster
 Posts: 53
 Joined: Sat Jul 29, 2006 7:33 am
 Location: (CSE,DU), Dhaka,Bangladesh
post reply of previous submission of 10948
A lot of thanks to rio. Now i get ac. Hope you will help me in future.
No venture no gain
with best regards

ishtiaq ahmed
with best regards

ishtiaq ahmed
:( WA
the same code i used for 543 (Goldbach) dunno watz wrong getting a WA
tQ
Code: Select all
#include<iostream>
using namespace std;
#define RANGE 1000010
#define sqrtRANGE 1010
int isprime[RANGE];
int prime[RANGE/10];
int primecount;
inline void primesieve()
{
unsigned int i,j;
for (i=0;i<RANGE;++i)
isprime[i]=i&1;
isprime[2]=1;
isprime[1]=0;
for (i=3;i<=sqrtRANGE;i+=2)
if (isprime[i])
for (j=i*i;j<RANGE;j+=i*2)
isprime[j]=0;
primecount=0;
for (i=0;i<RANGE;++i)
if (isprime[i])
{
isprime[i]=primecount;
prime[primecount++]=i;
}
// cout <<primecount<<endl;
}
int main()
{
primesieve();
int n,i,j,a,b;
while(cin>>n,n)
{
a=0;
b=n1;
while (!isprime[b]) b;
b=isprime[b];
while (prime[b]+prime[a]!=n && a<=b)
{
while (prime[b]+prime[a]>n)
{
b;
}
while (prime[b]+prime[a]<n)
{
a++;
}
if (prime[b]+prime[a]!=n) a++;
}
if (a<=b)
cout <<n<<":"<<endl<<prime[a]<<"+"<<prime[b]<<endl;
else
{
// cout <<n<<endl;
cout<<n<<":"<<endl<<"NO WAY!"<<endl; //never happens!
}
}
return 0;
}
Try this case:
Thanks
Keep posting
Sapnil
Code: Select all
Input:
1000000
999999
99999
1
2
3
11
121
100
Output:
1000000:
17+999983
999999:
NO WAY!
99999:
NO WAY!
1:
NO WAY!
2:
NO WAY!
3:
NO WAY!
11:
NO WAY!
121:
NO WAY!
100:
3+97
Keep posting
Sapnil