電大面向?qū)ο蟪绦蛟O(shè)計(jì)作業(yè)答案參考資料小抄(存儲(chǔ)版)
【正文】 Airlines to help other animals. Although there are only 13 episodes in this series, the content is very pact and attractive. The animation shows the preciousness of friendship and how people should be brave when facing difficulties. Even adults recalling this animation today can still feel touched by some scenes. Secrets of the Heavenly Book Secrets of the Heavenly Book, (Chinese: 天書奇談 ) also referred to as Legend of the Sealed Book or Tales about the Heavenly Book, was released in 1983. The film was produced with rigorous dubbing and fluid bination of music and vivid animations. The story is based on the classic literature Ping Yao Zhuan, meaning The Suppression of the Demons by Feng Menglong. Yuangong, the deacon, opened the shrine and exposed the holy book to the human world. He carved the book39。s trip to the West to take the Buddhistic sutra. They met a white bone evil, and the evil transformed human appearances three times to seduce the monk. Twice Monkey King recognized it and brought it down. The monk was unable to recognize the monster and expelled Sun Wukong. Xuan Zang was then captured by the monster. Fortunately Bajie, another apprentice of Xuan Zang, escaped and persuaded the Monkey King to e rescue the monk. Finally, Sun kills the evil and saves Xuan Zang. The outstanding animation has received a variety of awards, including the 6th Hundred Flowers Festival Award and the Chicago International Children39。s education. Proponents of the show claim that it is merely for entertainment. Effendi Effendi, meaning sir and teacher in Turkish, is the respectful name for people who own wisdom and knowledge. The hero39。 } 程序運(yùn)行結(jié)果 Date: Date: Date: 修改部分： class Date{… public:void Print()。 int GetYear(){return year。 ()。 寫出程序的運(yùn)行結(jié)果。 cout () . () . () endl。//取年份 int getmonth ()。 cout sum of array: sumofarray endl。 } 答： 語句號(hào) 1 2 3 4 5 對(duì) /錯(cuò) 錯(cuò) 對(duì) 錯(cuò) 錯(cuò) 對(duì) 三、 填空題 （ 20 分 ） 下面是一個(gè)求數(shù)組元素之和的程序。這些空間應(yīng)隨著對(duì)象的消亡而釋放掉，所以，需要在析構(gòu)函數(shù)中釋放這些空間。 （ E） top = pNew。 53. } 54. //模板類 Stack 的析構(gòu)函數(shù) ~Stack()的實(shí)現(xiàn) 55. templateCLASS TYPE 56. Stack::~Stack() 57. { 58. StackItem *p = top, *q。 41. top = topnextItem。 } 28. private: 29. StackItem * top。 15. StackItem * nextItem。 } 二、理解問答題： （ 20分） 下面的文件 是一個(gè)堆棧類模板 Stack 的完整實(shí)現(xiàn)。 } 4．寫一個(gè)函數(shù)，找出給定數(shù)組中具有最小值的元素。 head = newNode。 函數(shù)返回值：當(dāng)成功地插入新的節(jié)點(diǎn)時(shí)，函數(shù)返回指向新鏈表頭一節(jié)點(diǎn)的指針，否則，若不能申請(qǐng)到內(nèi)存空間，則返回 NULL。 else return sum(n1) ()*(1)n。 str[i] = 39。 函數(shù)參數(shù)： str 為所要處理的字符串； 函數(shù)返回值：所給字符串中數(shù)字字符的個(gè)數(shù) 答： int CalcCapital (char *str) { if (str == NULL) return 0。 函數(shù)的原型： int CalcCapital (char *str)。amp。 提示：你可以使用遞歸表達(dá)式： sum(n) = sum(n1) (1/n)*(1)n 答： float sum(int n) { if (n == 1) return 1。第二個(gè)參數(shù)newValue 為所給定的插入新節(jié)點(diǎn)的新數(shù)值。 //插入到空鏈表的表頭 else if(newValue=headValue){ newNodenext=head。 //將新節(jié)點(diǎn)插入 } } return head。
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