Số nguyên tố là thứ quá đỗi quen thuộc với những ai học toán, cũng như đối với các lập trình viên. Ai học giải thuật đều đã giải qua những bài về số nguyên tố, với độ khó dàn trải từ đơn giản tới phức tạp, từ vòng lặp thử tính chia hết tới sàng số nguyên tố Eratosthene…

Hôm nay mình rảnh thì có nghĩ ra một bài như thế này, các bạn xem qua nhé!

Đề bài: Cho hai số nguyên dương x và y (các test case đưa vào luôn đảm bảo x > y). Viết chương trình kiểm tra xem x^{2} – y^{2} có phải là số nguyên tố hay không.

Thoạt nhìn qua thì bài này khá đơn giản. Tính x^{2} – y^{2}, rồi kiểm tra. Nhưng khi x và y lớn, ví dụ cỡ 10^{18} thì cách này sẽ chạy quá thời gian.

Hãy quan sát biểu thức x^{2} – y^{2}. Nếu bạn đã học 7 hằng đẳng thức đáng nhớ, sẽ dễ thấy rằng x^{2} – y^{2} có thể được viết dưới dạng (x-y)(x+y). Vì x và y đều là các số nguyên dương, nên x ≥ 1 và y ≥ 1. Từ đó suy ra: (x+y) ≠ 1.

Xét biểu thức còn lại, x-y. Sẽ xảy ra hai trường hợp là (x-y) ≠ 1 và (x-y)=1. Với trường hợp đầu tiên, x^{2} – y^{2} sẽ không là số nguyên tố vì vi phạm định nghĩa (chỉ có hai ước là 1 và chính nó). Do đó với (x-y)=1, biểu thức ban đầu sẽ tương đương với chỉ x+y. Ta chỉ cần xét x+y là đủ. Sau đây là code miêu tả bài toán này.

bool primeSquare(long long a, long long b)
{
if (a-b != 1) return false;
long long num = a + b;
if (num % 2 == 0) return false;
for (long long i = 3; i*i <= num; i += 2)
{
if (num % i == 0) return false;
}
return true;
}

We all know a random number times 0 always ends up 0, but the division for 0 is impossible. Computer programs return numerous errors upon stumbling across this matter, math leaners try their best with conditions to avoid dividing by 0. Some guys on YouTube even tried to prove that 1 equals to 0 using this very infamous numerical expression.

But why is that? Let us take a very straightforward and intuitive example to explain why dividing by 0 is a meaningless act.

Say you have a delicious cake. Rashford and Pogba come to your house to eat it with you. In order to make everyone happy, the cake must be divided into 3 equal parts, so each one of you will have a third of that cake. In another scenario, it is the same cake, but you celebrate something alone and want to eat the whole thing. Simple. You divide the cake by 1, so it ends up being itself. Bon appetit!

But what happens when you attempt to divide the cake by 0? The cake just disappears. Just a second ago, a delicious piece of culinary art stays in front of you, but now it is in the void. The act itself gets ridiculous when you think about it. Where is the cake? Ask the divisor.

That is the answer for dividing by 0. You cannot destroy matters, because the act of dividing by 0 “destroys” matters and numbers. For those who studied physics, you must be familiar with the Conservation of mass: “mass can neither be created nor destroyed, although it may be rearranged in space, or the entities associated with it may be changed in form.”

You may be familiar with quick-sort, but when you stumble across this problem, it turns out to be a lot more complex. Since you need to determine the method that involves the fewest moves possible, you will have to take another approach. The given problem can be re-stated into this problem:

Problem statement

Hien is the class monitor and he wants his classmates to form a line, in which the height of every students is in ascending order. He needs to form that line by moving his classmates from the line to the start or end of it and it has to be a quick process, since Hien is very lazy and needs to play Age Of Empires right away. Write an algorithm to help him.

Input format

First line: n.

Second line: n numbers indicating the height of every student in the class, each seperated by a space.

Constraints

n ≤ 100; H[i] ≤ 100000 (H[i] is the height of an individual).

Output format

An integer indicating the fewest moves possible.

Sample input

4

2 1 3 5

Sample output

1

Explanation

The student with height 1 is moved to the start of the line.

Let’s not pay attention to the ‘fewest moves’ for a while. Normally, when you see these types of ‘moving’ elements to start or end of an array, you can take a look at a basic approach.

Let’s take the Sample input as an example. With the basic approach, we search for the smallest element in the array, which is now 1. After that, we move it to the far right of the array. Then, we search for the next smallest element, which is 2, and we keep doing it until 5 is moved to the far right of the array. We come to the conclusion that for this approach, the number of moves that are taken is exactly equal to the number of elements present in the array itself. But let’s have a closer look. We can see that 2, 3 and 5 are contiguous, meaning that the relative order between them is not changed at all when sorting is completed. So, we can know that in the required algorithm, we need to conserve the order of contiguous integers. That is when std::pair comes to use.

What is std::pair, exactly?

Std::pair is a pre-defined class in C++. A pair element is consisted of 2 other sub-elements, which can be classified as first and second. In the algorithm we are searching for, as stated earlier, we can utilise std::pair to get the job done by assigning the input elements to first and its index to second.

Get the job done

After that, quick-sort comes in handy. We then use it to sort the array in ascending order. Because std::pair is used, when sorting the elements, each index is carried along with the data. Then, we can just compare the indexes of every subsequent element. Job done!