[Data Structures and Algorithms - 1] Introduction & Mathematics

References:

1. Stanford University CS97SI by Jaehyun Park

2. Introduction to Algorithms

3. Kuangbin's ACM Template

4. Data Structures by Dayou Liu

5. Euler's Totient Function


Getting Started: 

1) What is a good algorithm?

The answer could be about correctness, time complexity, space complexity, readability, robustness, reusability, flexibility, etc.

 

However, in competitive programming, we care more about 

  • Correctness - It will result in Wrong Answer(WA)
  • Time complexity - It will result in Time Limit Exceeded(TLE)
  • Space complexity - It will result in Memory Limit Exceeded(MLE)

 

In algorithms contest, we need to pay attention to the time limit, memory limit, the range of input and output.

 

Example: A+B problem 

int x;
int y;
cin >> x >> y;
cout << x+y;

1+2 is ok

1+999999999999999 will result in overflow

 

2) How to prove correctness? 

  • Prove by contradiction
  • Prove by induction(Base case, inductive step)

Example: T(n) = T(n-1) + 1, T(1) = 0. Prove that T(n) = n - 1 for all n > 1 and n is an integer.

Proof:

(Base case) When n=1, T(1) = 1-1 = 0. It is correct.

(Inductive Step) Suppoer n = k, it is correct. T(k) = k - 1.

For n = k + 1, T(k+1) = T(k) + 1 = k - 1 + 1 = k. It is correct for n = k + 1. 

Therefore, the algorithm is correct for all n > 0 and n is an integer.

 

3) Big O Noatation

O(1) < O(log n) < O(n) < O(nlog n) < O($n^2$) < O($n^3$) < O($2^n$) 


 

1. Algebra 

1.1 Simple Algebra Formulas:

$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

$$\sum_{k=1}^n k^3 = (\sum k)^2= (\frac{n(n+1)}{2})^2$$

 

 

1.2 Fast Exponentiation

How to calculate $x^k$?

$x^k = x*x*x...x$

Notice that:

$x*x = x^2$

$x^2 * x^2 = x^4$

...

double pow (double x, int k) {
       if(k==0) return 1;
       if(k==1) return x;
       return k%2==0?pow(x,k/2)*pow(x,k/2):pow(x,k-1)*x;
}

(Important to consider special cases when you design an algorithm)

 1) k is 0 

 2) k is 1

 3) k is even and k is not 0

 4) k is odd and k is not 1

 

2. Number Theory

 2.1 Greatest Common Divisor(GCD)

gcd(x,y) - greatest integer divides both x and y. 

- gcd(a,b) = gcd(a, b-a)

- gcd(a, 0) = a

- gcd(a,b) is the smallest positive number in{$ax+by | x, y \in \mathbb{Z} $ } 

 

$x\equiv y\ (mod\ m) \Rightarrow a\%m=b\%m$ 

Properties: 
If $a_1 \equiv b_1(mod\ m), a_2 \equiv b_2(mod m)$, then:
$a_1 +a_2 \equiv b_1+ b_2(mod\ m)$
$a_1 -a_2 \equiv b_1- b_2(mod\ m)$
$a_1 *a_2 \equiv b_1* b_2(mod\ m)$

 

 

  • Euclidean algorithm
int gcd(int a, int b) {
    while(b) {int r = a%b; a = b; b = r;}
    return a;
}

 

 

  • Extended Euclidean algorithm

Problem: Given a,b,c. Find integer solution x,y for ax+by=c.

If c % gcd(a,b) = 0, there are infinite many solutions. Otherwise, there is no solution.

long long extended_gcd(long long a, long long b, long long &x, long long &y) {
    if(a==0 && b==0) return -1;
    if(b==0) {x=1,y=0; return a;}
    long long d=extended_gcd(b, a%b, y, x);
    y -= a/b*x;
    return d;
}

 

 

2.2 Prime Numbers

  • For any N$\in \mathbb{Z} $,there is $N=p_1^{e1}p^{e2}_2...p^{er}_r$. And $p_1,p_2, ..., p_r$ are prime numbers. The number of factors for N is $(e1+1)(e2+1)...(er+1)$.

 

  • Sieve's code
void getPrime(int n) { 
    int i, j;
    bool flag[n + 1];
    int prime[n + 1];
    memset(flag, true, sizeof(flag)); // suppose they are all prime numbers
    int count = 0; // the number of prime numbers
    for(i = 2; i <= n; ++i) {
        if(flag[i]) prime[++count] = i;
        for(j = 1; j <= count && i*prime[j] <= n; j++) {
            flag[i*prime[j]] = false;
            if(i%prime[j] == 0) break;
        }
    }
}

 

 

2.3 Bionomial Coefficients

${n}\choose{k} $= $\frac{n(n-1)...(n-k+1)}{k!}$

Use when both n and k are small. Overflow risk.

 

 

2.4 Euler's Function

$n=p_1^{n_1} * p_2^{n_2} * ... p_k^{n_k}$
$\varphi(x) = x(1-\frac{1}{p_1})(1-\frac{1}{p_2})...(1-\frac{1}{p_k}) $

int getPhi(int x)
{
    float ans = x;   
    for (int p=2; p*p<=n; ++p){
        if (x % p == 0){
            while (x % p == 0)
                x /= p;
            ans*=(1.0-(1.0/p));
        }
    }
    if (x > 1)
        ans*=(1.0-(1.0/x));
    return (int)ans;
}

 

 


 

Practice Problems: (HDU, POJ, UVa - https://vjudge.net/ ; LeetCode - leetcode.com)

POJ 1061, 1142, 2262, 2407, 1811, 2447

HDU 1060, 1124, 1299, 1452, 2608, 1014, 1019, 1108, 4651

LeetCode 204

UVa 294

 

 

 

posted @ 2018-01-26 15:00 rgvb178 阅读(...) 评论(...) 编辑 收藏