Main Page

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Mod properties

(a+ b) mod m = (a mod m+ b mod m) mod m (a− b) mod m = (a mod m− b mod m) mod m (a· b) mod m = (a mod m· b mod m) mod m

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Comparing double values

if (abs(a-b) < 1e-9) { // a and b are equal }

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Find divisors of a number in $$O( \sqrt {n} )$$

for (int i = 2; i * 1ll * i <= x; ++i) {
    if (x % i == 0) {
        dd.push_back(i);
        if (i != x / i) {
            dd.push_back(x / i);
        }
    }
}
            

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When asked how to make two numbers equal by series of divisions Represent those numbers in terms of powers of those numbers. for eg allowed operations are /2,/3,/5 then A=x*2^a * 3^b * 5^c B=y*2^a2 * 3^b2 * 5^c2 if x is same as y it can be made equal

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Goldbach's Conjecture

It says any number can be expressed as a sum of 2 prime numbers.

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Fast nCr

Use this piece of code to calculate nCr faster.

                    
        ll NcR(ll n, ll r) 
        {  
            ll p = 1, k = 1;  
            if (n - r < r) 
                r = n - r; 
        
            if (r != 0) { 
                while (r) { 
                    p *= n; 
                    k *= r; 
                    ll m = __gcd(p, k); 
                    p /= m; 
                    k /= m; 
                    n--; 
                    r--; 
                } 
            } 
        
            else
                p = 1; 

            return p;
        } 
                    
                

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Antidifference Function

Difference of a function is Δf(x) = f(x+1) - f(x) = F(x) Anti difference of a function is Δ^−1F(x) = f(x). “The antidifference of a function f(x) is the function g(x) such that f(x) = g(x + 1) − g(x). So, if we have a summation of f(x), it can be simplified by the use of its antidifference in the following way: f(k) + f(k + 1) + f(k + 2) + . . . + f(k + n) = = g(k + 1) − g(k) + g(k + 2) − g(k + 1) + g(k + 3) − g(k + 2)+ . . . +g(k + n + 1) − g(k + n) = = g(k + n + 1) − g(k)