### PROBLEM LINK:

### DIFFICULTY:

EASY

### 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®] 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* = 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 ",ans); } }`

**Approach 2:**

This approach is taken by more coders including tester.

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). See tester’s solution for working code.

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution will be uploaded soon

Tester’s solution can be found here and here