欧美一级a免费放视频,欧美一级a免费放视频_丰满年轻岳欲乱中文字幕电影_欧美成人性一区二区三区_av不卡网站,99久久精品产品给合免费视频,色综合黑人无码另类字幕,特级免费黄片,看黃色录像片,色色资源站无码AV网址,暖暖 免费 日本 在线播放,欧美com

合肥生活安徽新聞合肥交通合肥房產(chǎn)生活服務(wù)合肥教育合肥招聘合肥旅游文化藝術(shù)合肥美食合肥地圖合肥社保合肥醫(yī)院企業(yè)服務(wù)合肥法律

代寫 tic-tac-toe game 、代做Python/Java程序語言

時間:2024-06-09  來源:合肥網(wǎng)hfw.cc  作者:hfw.cc 我要糾錯



Homework 4 Revision 1
Due: June 6, 2024; late due date is June 11
Points: 100
1. In problem 1, if the square is occupied, you need to give the error message. “%c has played %d,%d\n” (where “%c” is either “X” or “O”, whichever is already in the square, and “%d,%d” are the co-ordinates of the occupied square).
2. In problem 1, the dimensions of the “O” was added; it is to be 5 × 5 centered in the square.
3. For all problems, the exit status code is 0.

(25 points) Enhance the tic-tac-toe game so 2 people can play. To do this, you need to add an“O” that, like the “X”, is drawn in the square. Then prompt the user for a square identifier, and alternate between drawing “X”s and “O”s at those locations on the board. The first move is for “X”. You need to detect and reject when a user plays a square that is already taken. Stop after the board is full (that is, 9 plays). You do not have to worry about who wins.
The “O” is to be 5 × 5, centered in the middle of the square.

Here are the messages your program should print to the standard output:

The tic-tac-toe board, with “X”s and “O”s as appropriate;
When it is “X”’s turn, print “X’s turn > ” (note the space after the “>”); and
When it is “O”’s turn, print “O’s turn > ” (again, note the space after the “>”).
Here are the error messages; all are to be printed on the standard error:

When the user enters only 1 co-ordinate: “Need 2 co-ordinates\n” (the ‘\n’ is a newline);
When there is an illegal character in the input: “Illegal character in input "%c"\n” (the “%c” is to print the offending character); and
When the square is already occupied: “%c has played %d,%d\n” (where “%c” is either “X” or “O”, whichever is already in the square, and “%d,%d” are the co-ordinates of the occupied square); and
When an invalid set of co-ordinates are entered: “%d,%d” is not a valid square; the numbers must be between 1 and 3 inclusive\n” (each %d is one of the invalid numbers).
If the program reads an end of file at the prompt, print a newline and quit.

The program should exit with an exit status code of 0.

Please call your program ttt4a.c and submit it through Gradescope. A sample executable, used to generate the Gradescope validation outputs, is available on the CSIF at /home/bishop/hw4/ttt4a.

(25 points) Now enhance what you did for question 1. Have your program determine when the game is over — that is, there is a winner (three in a row, column, or diagonal) or a tie (no winner and all 9 squares on the board are full). Print the results on the standard output:
If X wins, print “Game over! X won!\n”;
If O wins, print “Game over! O won!\n”; or
If X wins, print “Game over! It’s a tie!\n”.
Please call your program ttt4b.c and submit it through Gradescope. A sample executable, used to generate the Gradescope validation outputs, is available on the CSIF at /home/bishop/hw4/ttt4b.

(50 points) The birthday problem asks how many people must be in a room so that the probability of two of them having the same birthday is 0.5. This problem has you explore it by simulation. Basically, you will create a series of lists of random numbers of length n = 2, …, and look for duplicates. You will do this 5000 times for each length. For each length, count the number of lists with at least 1 duplicate number; then divide that number by 5000. That is the (simulated) probability that a list of n generated numbers has at least one duplicate. As the random numbers you generate are between 1 and 365 (each one corresponding to a day of the year), this simulates the birthday problem.
Now, breathe deeply and calm down. We will do this in steps; you only have to turn the final program in to Canvas (not Gradescope).

First, detecting duplicates. Write a function called hasduplicates(bday) that takes an array bday and returns 1 if it contains a duplicate element, and 0 if it does not. For example, if bday is

int bday[] = { 1, 2, 3, 4, 5, 5, 2 };
then hasduplicates(bday) returns 1 and if

int bday[] = { 1, 2, 3, 4, 5, 6, 7 };
then hasduplicates(bday) returns 0.
Now, deal with one set of birthdays. Write a function called onetest(count) that generates a list of count random integers between 1 and 365 inclusive, and returns 1 if it contains a duplicate element, and 0 if it does not. Please use the function hasduplicates(bday) to test for duplicates.
Now for the probability for count people. Write a function probab(count, num) that runs num tests of count people, and counts the number of tests with duplicates. It returns the fraction of the tests with duplicates; that is, the number of duplicates divided by num.
Now for the demonstration. Start with 2 people, and begin adding people until the probability of that many people having two people with a birthday in common is over 0.5. (In other words, start with a list of 2 elements, and increase the number of elements in the list until the simulation shows a probability of 0.5 that a number in the list is duplicated.) Print each probability; your output should look like this:

For  2 people, the probability of 2 birthdays in common is 0.001400
For  3 people, the probability of 2 birthdays in common is 0.006600
For  4 people, the probability of 2 birthdays in common is 0.015200
For  5 people, the probability of 2 birthdays in common is 0.025400
For  6 people, the probability of 2 birthdays in common is 0.041400
For  7 people, the probability of 2 birthdays in common is 0.053000
For  8 people, the probability of 2 birthdays in common is 0.082000
For  9 people, the probability of 2 birthdays in common is 0.092200
For 10 people, the probability of 2 birthdays in common is 0.121800
Hint: Don’t be surprised if your probabilities are slightly different than the ones shown in the sample output. As randomness is involved, it is very unlikely your numbers will match the ones shown here.
To turn in: Please call your program bday.c and submit it through Canvas (not Gradescope). A sample executable is available on the CSIF at /home/bishop/hw4/ttt4b.
請加QQ:99515681  郵箱:[email protected]   WX:codinghelp























 

掃一掃在手機打開當前頁
  • 上一篇:菲律賓9A簽證有哪些(9A簽證詳細介紹)
  • 下一篇:菲律賓旅游簽證回國流程(回國不能帶什么東西)
  • 無相關(guān)信息
    合肥生活資訊

    合肥圖文信息
    出評 開團工具
    出評 開團工具
    挖掘機濾芯提升發(fā)動機性能
    挖掘機濾芯提升發(fā)動機性能
    戴納斯帝壁掛爐全國售后服務(wù)電話24小時官網(wǎng)400(全國服務(wù)熱線)
    戴納斯帝壁掛爐全國售后服務(wù)電話24小時官網(wǎng)
    菲斯曼壁掛爐全國統(tǒng)一400售后維修服務(wù)電話24小時服務(wù)熱線
    菲斯曼壁掛爐全國統(tǒng)一400售后維修服務(wù)電話2
    美的熱水器售后服務(wù)技術(shù)咨詢電話全國24小時客服熱線
    美的熱水器售后服務(wù)技術(shù)咨詢電話全國24小時
    海信羅馬假日洗衣機亮相AWE  復古美學與現(xiàn)代科技完美結(jié)合
    海信羅馬假日洗衣機亮相AWE 復古美學與現(xiàn)代
    合肥機場巴士4號線
    合肥機場巴士4號線
    合肥機場巴士3號線
    合肥機場巴士3號線
  • 上海廠房出租 短信驗證碼 酒店vi設(shè)計