Shulou(Shulou.com)06/02 Report--

This article will explain in detail how to use C language to realize the reproduction and spread of bacteria. The editor thinks it is very practical, so I share it with you as a reference. I hope you can get something after reading this article.

one。 Introduction

This is the first article I wrote in csdn ~ as a beginner in C, I came across a question about the reproduction and spread of bacteria yesterday. I thought it was very interesting. After thinking about it all night, I finally came up with the solution.

two。 Details of the topic the reproduction and spread of bacteria

Total time limit: 1000ms memory limit: 65536kB

Description:

In a square petri dish with a side length of 9, there are m bacteria in the center. Suppose that the life span of bacteria is only one day, but 10 offspring can be produced every day, and two of the 10 offspring are distributed in the original cells, and the rest are evenly distributed in the eight adjacent cells around them. The distribution of bacteria in petri dish was studied after n (1 ≤ n ≤ 4) days.

Enter:

Enter two integers, the first integer m for the number of bacteria in the center (2 ≤ m ≤ 30), and the second integer n for the number of days elapsed (1 ≤ n ≤ 4).

Output:

Outputs a matrix of nine rows and nine columns of integers, with each row of integers separated by spaces. The whole matrix represents the distribution of bacteria on the petri dish after n days.

Sample input

twenty-one

Sample output

0 0 0

0 0 0

0 0 0

0 0 0 2 2 2 0 0 0

0 0 0 2 4 2 0 0 0

0 0 0 2 2 2 0 0 0

0 0 0

0 0 0

0 0 0

three。 Ideas and code ideas

My idea is that this problem needs to define two two-dimensional arrays, one to store the previous day's data (old), and one to calculate and save the day's data (new). The data of this day can be saved to the old data for the next day's calculation (recursion).

First of all, each bacteria in its original position can have two cubs in its original position and leave a newborn in each of the eight squares around it. And then unfortunately apoptosis.

In this way, we first save the data of the previous day in the old data matrix, and then multiply each data of the previous day by two into the matrix of the new data. Then do another cycle to judge that if the data at a certain position in the old data matrix is not zero, the data is added to all eight cells around the corresponding position of the new data matrix, which is equivalent to the spread of bacteria in the surrounding cells. In this way, we complete the recursive process.

Note that this recursion can only last four days (loop four times), otherwise a stack error may occur.

The code # includeint main () {int old [9] [9], birth [9] [9]; / / the old medium and the new medium for recursive int m, ntternal / initial number of central bacteria and recursive days int I, j, p; scanf ("% d% d", & m, & n); for (I = 0% I < 9) Birth +) for (j = 0bot j < 9x j +) {birth [I] [j] = 0politics / initialization of variables} birth [4] [4] = m bank / central bacteria have grown for (p = 1bot p

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