# PROBLEM LINK

# DIFFICULTY

MEDIUM

# PREREQUISITES

Prime Factorization, Sieve

# PROBLEM

Find the number of integers in the given range which have prime number of divisors.

# QUICK EXPLANATION

Although L and R can be as high as 10^12 but R-L is <= 10^6, this allows us to iterate through the complete range and count the number of divisors of each number in the range.

# EXPLANATION

**Approach 1:**

If a number N is divisible by D then either it has 2 factors(D and N/D) or it is square of D. Another thing to note is that if a number has 2 divisors x and y such that x < y, then x < sqrt(N) and y > sqrt(N). Using these 2 facts we can iterate through all the numbers in range [1, sqrt(R)] and count the number of factors for each of the number.

Working commented solution follows:

```
bool isPrime[10000];
void init() {
// Since range is very small so not used Sieve
for (int i = 2; i < sizeof(isPrime); ++i) {
int j = 2;
for (; j * j <= i; ++j) {
if (i % j == 0) {
break;
}
}
if (j * j > i)
isPrime[i] = true;
}
}
main() {
init();
int testCases, divisors[1000005];
scanf("%d", &testCases);
while(testCases--) {
long long L, R;
scanf("%lld%lld", &L, &R);
for(long long i=L; i<=R; i++)
divisors[i-L] = 0; //Initialize divisors of all numbers to 0
for(long long i=1; i*i <= R; i++) { // Iterate through 1 to sqrt(R)
long long square = i*i; // j starts with first number in range [L, R] which is multiple of i
for(long long j=( (L-1) / i + 1) * i; j <= R; j += i) { // Factors under consideration are i and j / i
if (j < square) continue; // Already counted because of 2 in else condition below
if( square == j ) // Just 1 factor
divisors[j-L] += 1;
else divisors[j-L] += 2; // Counting factors i and j / i
}
}
int ans = 0;
for(long long i = L; i <= R; i++)
if(isPrime[divisors[i-L]])
ans++;
printf("%d\n",ans);
}
}
```

**Approach 2:**

This approach is taken by more coders.

This is based on prime factorization of a number and then counting the total number of divisors.

Lets take an example:

Consider prime factors breakdown of 18 (2^1 * 3^2)

so total number of factors are (1+1) * (2 + 1) = 6 [Factors: 1, 2, 3, 6, 9, 18].

Similarly if a number is p1^x1 * p2^x2 * p3^x3 * … then total number of factors is (x1+1) * (x2+1) * (x3+1) * …

Now if we generate all primes <= 10^6 using sieve and later use them to find out how many times it goes into each number in the given range, we know the number of divisors composed of primes <= 10^6.

We may have missed 1 prime number > 10^6 which might have fallen in the range as range goes as high as 10^12. But we are sure that there is only 1 such prime! So if we detect this case we can simply multiply our number of factors calculated so far by (1+1).