Programming masters, please enter 50, ask for programming gods

The first question: there are 9x8x7x6x5x4x3x2x1 cases, just write in a loop, and recursively can also be.

Second question:

Some people may think that this is actually quite simple, 55x54 directly....x1 will do the trick. In fact, people who think so don't see the real point of this question, that is, the representation of large numbers in the computer.

There is no common data type in a computer that can represent "55! "That's a lot.

Therefore, you need to construct a new data structure by yourself, and then write a matching algorithm to solve the problem!

How to write? I'm lazy. And this thing is really complicated to write, so I found you a material "Big Numbers".

This article is quite well written, it is specially trained by ACM, and it is enough for you

The first question is 9x8x7x6x5x4x3x2x1

The second 55x54....x1

The first problem and the second problem are really an algorithm.

Let x, y, and z be the number of roosters, hens, and chicks to buy, respectively.

Do it with MATLAB:

x y]=meshgrid(0:20,0:33);

z=100-5*x-3*y;

The three two-dimensional arrays of x, y, and z are obtained, and the elements with the same subscript are all solutions of the equation;

At the same time, x, y, and z should all be greater than 0.

n%10 represents the remainder of n divided by 10, which is the single digit of n.

n 10)%10 is the number of tens of digits.

This person who has studied C will, it's easy.

Go to the NetEase open class to find information and study.

/2.Minute conversion.

int h = 560/60;

int m = 560-60*h;

hours"+m+"minutes");

3.Single, ten, and hundred-digit sums.

Please enter a three-digit number:");

scanner sc = new scanner(;

int num = ;

int sum = num%10+num%100/10+num/100;

And yes:"+sum);

4.Swap variables.

Please enter two variables:");

scanner scanner = new scanner(;

int num1 = ;

int num2 = ;

The order before the exchange is:"+num1+" "+num2);

int num3 = num1;

num1 = num2;

num2 = num3;

The order after the exchange is:"+num1+" "+num2);