# https://www.codechef.com/LRNDSA07/problems/KPRIME

My code is giving TLE can anyone optimize my code.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define vi vector<int>
#define ff first
#define ss second
#define ins insert
#define endl "\n"
#define pie_op ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define tc int t;cin>>t;while(t--)
#define pb push_back

const int MOD=1e9+7;

vi pr(100001, 0);
int dp[6][100001]={0};

void seive(){
for(int i=1;i<=100000;i++){
int n=i;
for(int j=2;j<=n;j++){
if(n%j==0){
while(n%j==0)
n/=j;

pr[i]=1+pr[n];
n=1;
break;
}
}
if(n>1){
pr[i]=1;
}
}
for(int i=1;i<=100000;i++){
for(int j=1;j<=5;j++){
if(pr[i]==j){
dp[j][i]=dp[j][i-1]+1;
}
else
dp[j][i]=dp[j][i-1];
}
}
}

int main()
{
pie_op;
seive();
tc{
int ans=0;
int a,b,k;cin>>a>>b>>k;
cout<<dp[k][b]-dp[k][a-1]<<endl;
}
return 0;
}

Change the code to calculate the number of unique prime factors of a number.

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
const int nax = 2e6 + 9;
const int MOD = 1e9 + 7;
int prime[nax];
void sieve() {
for (int i = 2; i < nax; ++i) {
if(prime[i] > 0) continue;
for (int j = 2; (i * j) < nax; ++j)
++prime[i * j];
}
}
void inline test_cases() {
int A, B, K, res = 0;
cin >> A >> B >> K;
for (int i = A; i <= B; ++i)
if(prime[i] == K) ++res;

cout << res << endl;
}

signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// #ifndef ONLINE_JUDGE
//    freopen("input.txt", "r", stdin);
// #endif
sieve();
int t = 1;
// cin >> t;
while(t--) {
test_cases();
}
return 0;
}

while running the sieve, you can mark the prime factor of each number.

it is not getting accepted. and i think this will also give TLE

I just submitted the code.
https://www.codechef.com/viewsolution/45739460