Just estimate each number, if it is lower than 168, then this underpass is unable to be passed. If no number is less than 168 ,then there is no crash.
Turn the pseudo code in to real code. And an important thing is that the value of i and j is not fixed that i < j. So we have to judge the value of i j.
The range of the number is very small, only 0 to 9, store all the factorial of n, and then calculate then one by one.
This is an easy problem and you just need to stimulate the hole process.
The easiest problem
First, you get the fist pair of numbers. And then find a pair of numbers that is totally different with the first pair.
To make the most product, you just need to change all numbers to negative first. And then find the smallest number and make it positive.
This problem has a regular formula. First you count the outermost layer has how many squares. Then the second outermost layer. Until the second smallest layer.
The algorithm of the problem is BFS or DFS. I use BFS. Traverse all the points by layer and count it's alongside cats.
The answer of the problem has a regular pattern. For example, there is a sequence A and for number An in A, the number that is bigger than An has two parts
At first I didn't understand the meaning of the problem. It means 1 is the root and the distance between the points on the first layer with the root is 1.
This is to find how many strings that starts with A and ends with B and the number of AB inside the string.
The max number = max(max number, answer).
The main mission of this problem is to count each players number of right questions.
Seeing the operation, we can easily think of Euclidean algorithm which is the way to calculate GCD.
