Codeforces round #147 div 2 - Problem C

/*
This program solve Problem C codeforces round #147 div 2
Date: 26/10/2012
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*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
using namespace std;
const int maxn = 10000010;
//------------------------------------
bool isprime[maxn];
int nPrime[maxn];
int a,b,k;
//------------------------------------
/*Sieve for all prime from 1 to 1000000
 nPrime[i] indicates the number of prime numbers from 1 to i
 isprime[i] indicates whether i is prime or not
*/
void sieve() {
 int i,j;
 i=2;
 memset(isprime,true,sizeof(isprime));
 isprime[1]=false;
 nPrime[1]=0;
 nPrime[2]=1;
 while (i<=int(sqrt(maxn)))
 {
  if (isprime[i])
  {
   j=i+i;
   while (j<=maxn) 
   {
    isprime[j]=false;
    j+=i;
   }
   i++;
  }
  else i++;
 }
 for (int i=3; i<=maxn; i++)
  if (isprime[i]) nPrime[i]=nPrime[i-1]+1;
  else nPrime[i]=nPrime[i-1];
}
//------------------------------------
/* Check if the value of l satisfies the problem condition */
bool ok(int l) { 
 int eps;
 for (int i=a; i<=b-l+1; i++)
 {
  eps=0;
  if (isprime[i]) eps++;
  if (nPrime[i+l-1]-nPrime[i]+eps<k) return(false);
 }
 return(true);
}
//------------------------------------
/* Search for the smallest value satisfies from 1 to b-a+1*/
int search(int a, int b) {
 int l,r,m,eps;
 l=a; r=b; 
 do
 {
  m=(l+r)/2;    // Take middle value
  if (ok(m)) r=m;   // If ok, search for the lower bound
  else l=m;    // If not, it must be somewhere in the upper bound
 } while (r-l>1);
 if (ok(l)) return(l);  // Lower bound first...
 else if (ok(r)) return(r); // Then upper bound
 return(-1);     // Cant find anything
}
//------------------------------------
int main() {
 sieve();
 cin >> a >> b >> k;
 cout << search(1,b-a+1);
 return 0;
}
//------------------------------------