I am new to this forum and not well aware of protocols of this forum so pardon me for my ignorance. My question is related to spoj problem https://www.spoj.pl/problems/KPRIMES2/. I am getting TIME LIMIT EXCEED for this problem.I think the bottleneck of this program is generating 10^9.Could some one suggest how to improve this sieve , faster way to generate prime or how to solve this problem. Here is sketch of my algorithm
This program generates all the primes of form 2k+1 and encoded these primes into 32 bit integers of array a[i] in which unset bit represents primes.a[0] encodes 3,5,7.......65.a[1] encodes 67 onwards and so on. I have taken a auxiliary array bitcnt[] , in which bitcnt[i] stores sum of unset bits of a[0], a[1],.........a[i]. I used bitcnt for binary search and find the position of kth number.Here is bit explanation of functions. prime() function generated primes and i encoded the primes onto bits of number[32 bit unsigned integer]. bitcnt array stores sum of unset bits of array a for binary search purpose. bsearchupper(int m) return index of bitcnt in which m lie. Finally in main function , i am storing how many primes are upto upperbound of m and started decreasing value till i got K. Thank you.
Edit:Problem statement from SPOJ
Input
An integer stating the number of queries Q(equal to 100000), and Q lines follow, each containing one integer K between 1 and 50000000 inclusive.
Output
Q lines with the answer of each query: the Kth prime number.
Example
Input: 8 1 10 100 1000 10000 100000 1000000 10000000
Output: 2 29 541 7919 104729 1299709 15485863 179424673
#include<cstdio>
#include<vector>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<ctime>
#define Lim 1000000000
using namespace std;
unsigned int a[(Lim>>6)+10],bitcnt[(Lim>>6)+10];
int bound;
void prime()
{
int p_1,q_1,p_2,q_2,Ub=sqrt(Lim*1.0);
for(int i=3;i<=Ub;i+=2)
{
p_1=(i-3)>>6,q_1=((i-3)>>1)&31;
if(!(a[p_1] & (1L<<q_1)))
for(int j=i*i;j<Lim;j+=i)
if(j&1)
{
p_2=(j-3)>>6,q_2=((j-3)>>1)&31;
a[p_2]|=(1L<<q_2);
}
}
int cnt=0;bound=0;
for(int i=0; i<=((Lim>>6)-1);i++)
{
//p_1=(i-3)>>6,q_1=((i-3)>>1)&31;
cnt+=__builtin_popcount(~a[i]);
bitcnt[bound++]=cnt;
//cout<<bound-1<<"---"<<bitcnt[bound-1]<<endl;
}
//cout<<cnt<<endl;
}
int bsearchupper(int m)
{
int lo=0,hi=bound,mid;
while(lo<hi)
{
mid=lo+((hi-lo)>>1);
if(bitcnt[mid]<=m)lo=mid+1;
else hi=mid;
}
//cout<<"lo= "<<lo<<" mid= "<<mid<<" hi= "<<hi<<endl;
return lo;
}
int main()
{
//clock_t start,end;
//start=clock();
prime();
int t,k,c,ret,w;
for(scanf("%d",&t);t>0;t--)
{
scanf("%d",&k);
if(k==1) {cout<<"2"<<endl;continue;}
k=k-2;
c=bsearchupper(k);
ret=bitcnt[c],w=32*(c+1);
for(int i=31;i>=0;i--)
{
if(!(a[c] & (1L<<i)))
{
ret--;
if(ret==k) printf("%d\n",3+(w-1)*2);
}
w--;
}
}
//end=clock();
//cout<<((end-start)/(double)CLOCKS_PER_SEC)<<endl;
}
