They re going to be able to explain a little bit of the process

Record back a record set newdyn, if the record set record is empty, it is required to enter data, and then from whth to end with** segment is to establish a data view according to the records in the database, the program can basically be copied, that is, you need to build a database yourself, which involves some specific details of the above things may need to be modified, depending on the situation. Hehe. If there's anything you don't understand, you can ask me.

The number of recursive calls was exceeded.

You can try increasing the number of recursions allowed, for example:

set(0,'recursionlimit', 1000) to see if it can run.

Possible reasons for this issue:

1. The recursive export conditions are not met due to the difference in software versions;

2. There is a random factor in the program, and the results may be different for different times.

The specific need to look at ** can be judged.

The first one: only Space and Enter can end the string, so A:11Space and 12 Enter are two.

b:11,12 is a string c:11,carriage return 12 carriage return is also two, so b is wrong ac right.

The second one: this question should be wrong.

That's how it should be.

2d is formatted input, meaning that only 2 int characters can be entered, while float will be a decimal point with no one digit after it, so j=55 y= name="777abc"

Some things will be known for themselves.

Obviously, this is the sort of choice. Each time, the largest platoon will be chosen to be in front.

Let's assume that the first number in the array is the largest, and assign it to q, and when you find that there is a larger number than this number, then assign this larger number to q, and after comparing it with all the numbers that follow, q is the largest number.

If a number greater than q is not found in the subsequent comparison, i.e., in the case of p=i, then there is no need for swapping.

If there is a larger number after it, i.e. if(q, then I have to swap the two numbers, and swap the largest number in front.

Finally, the order is formed from largest to smallest.

"If i is not equal to p, swap the values of a(p) and a(i) - what does this mean".

According to the results of a cycle, the maximum values will be arranged in order from largest to smallest.

When the data is the largest in the data after the position, there is no need to change the position of the data.

Otherwise, swap that maximum to the current position.

It is the idea of sorting, which should be swap sorting.

The result of i%2 is only 0 and 1, so the result of b---is only -1 and 0, and the meaning of the while loop is: only when b = 1, k++, so this program is used to calculate the number of numbers that are not divisible by 2 from 1 to 5, and then output the value of b during the last trip of the for loop, when i = 5, b = 1, and then b --, is equal to 0, so the final output result is 3, 0

<> "See if it's useful to match the bad reputation, written with C++.