# Why we prefer a check such as i*i <= n instead of i <= sqrt(n)?

Because we incur an extra log(n) cost when performing a sqrt function. Using the former variant means no cost at all.

# Global versus method local declaration of an array

In a method local array, the max size of an array is less than what could have been defined globally.

# Problem solving technique

For problems that involves factors etc. try out creative ways to store meta data in the sieve array itself. For example,

1. Sieve array can store number of times a prime say 7 is multiplied for a given large number n during the precomputation step. Refer L4 of Str lecture - the last problem discussed.
2. Store backpointers for all numbers 1 through n as to which prime number was used to mark this number as a composite. For this the array need to be initialized with the numbers themselves (e.g. a = 5, a = 6 and so on because doing so with also cause the primes to record the fact that they themselves are their factors). Refer L4 of Str lecture 25:00.
3. Fact: For a given number n, primes <= sqrt(n) are enough to generate all prime factors. For example, for 57 (3 * 19), 3 is enough to generate the remaining prime factors.

# Modulus function snippets

``````#define ll long long int

int mod = 1e9 + 7;

inline ll add(ll a, ll b) {
ll num = a + b;
if (num >= mod) {
num -= mod;
} else if (num < 0) {
num += mod;
}
return num;
}

inline ll sub(int a, int b) {
ll num = a - b;
if (num < 0) {
num += mod;
}
else if (num >= mod) {
num -= mod;
}
return num;
}

inline int mul(int a, int b) {
return (int) ((long long) a * b % mod);
}

inline int power(int a, long long b) {
int res = 1;
while (b > 0) {
if (b & 1) {
res = mul(res, a);
}
a = mul(a, a);
b >>= 1;
}
return res;
}

inline int inv(int a) {
a %= mod;
if (a < 0) a += mod;
int b = mod, u = 0, v = 1;
while (a) {
int t = b / a;
b -= t * a; swap(a, b);
u -= t * v; swap(u, v);
}
assert(b == 1);
if (u < 0) u += mod;
return u;
}

int fact[N];

void pre() {
fact = 1;
for(int i = 1; i < N; i++)
fact[i] = mul(fact[i - 1], i);
return;
}
``````

# Sieve of Eratothenes

``````int n = 200;
bool prime;

void SieveOfEratosthenes(int n)
{
memset(prime, true, sizeof(prime));

for (int p = 2; p * p <= n; p++) {
if (prime[p]) {
for (int i = p * p; i <= n; i += p) {
prime[i] = false;
}
}
}

/*for (int p = 2; p <= n; p++) {
if (prime[p]) {
cout << p << " ";
}
}*/
}
``````