Problem A 複製人來了！

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轉殖技術飛速發展，複製人已經成為現實了！！所以，現在由你來編寫乙個person類，來模擬其中的轉殖過程。這個類具有2個屬性：name——姓名（char*型別），和age——年齡（int型別）。

該類具有無參建構函式（人名為「no name」，年齡是0）、帶引數建構函式、拷貝建構函式以及析構函式外，還有以下3個成員函式：

1. void person::showperson()：按照指定格式顯示人的資訊。

2. person& person::setname(char *)：設定人的姓名。

3. person& person::setage(int)：設定人的年齡。

輸入分多行，第一行是乙個正整數n，表示其後有n行輸入。每行分兩部分：第一部分是乙個沒有空白符的字串，表示乙個人的姓名；第二部分是乙個正整數，表示人的年齡。

呃~比較複雜，見樣例吧！注意：要根據樣例編寫相應函式中的輸出語句，注意格式哦！

3zhang 20li 18zhao 99

a person whose name is "no name" and age is 0 is created!a person whose name is "tom" and age is 16 is created!a person whose name is "tom" and age is 16 is cloned!a person whose name is "zhang" and age is 20 is created!this person is "zhang" whose age is 20.a person whose name is "zhang" and age is 20 is erased!a person whose name is "li" and age is 18 is created!this person is "li" whose age is 18.a person whose name is "li" and age is 18 is erased!a person whose name is "zhao" and age is 99 is created!this person is "zhao" whose age is 99.a person whose name is "zhao" and age is 99 is erased!this person is "zhao" whose age is 18.this person is "no name" whose age is 0.a person whose name is "zhao" and age is 18 is erased!a person whose name is "tom" and age is 16 is erased!a person whose name is "no name" and age is 0 is erased!

注意：輸出中有「」！ [

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#include

usingnamespacestd;

classperson

person(string n):name(n),age(0)

person(string s,inta)

person(constperson & p)

~person()

voidshowperson()

person& setname(string nn)

person& setage(intaa)

};

intmain()

anothertom.setname(str).setage(18);

anothertom.showperson();

noname.showperson();

return0;

}

Problem A 複製人來了！

Problem A 複製人來了！

複製人來了！

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