基本程式設計筆試題
阿新 • • 發佈：2019-01-10
1.定義一個“資料型別”datatype,能處理字元型，整形，浮點型三種資料型別，給出建構函式。
#include"stdafx.h" #include<iostream> using namespace std; class datatype { enum { Inter, FLoat, Char }vartype; union { char c; int i; float f; }; public: datatype(int ii) { vartype = Inter; i = ii; } datatype(char cc) { vartype = Char; c = cc; } datatype(float ff) { vartype = FLoat; f = ff; } void print(); }; void datatype::print() { switch (vartype) { case Inter: cout << "整形：" << i << endl; break; case Char: cout << "字元型：" << c << endl; break; case FLoat: cout << "浮點型：" << f << endl; break; default: break; } } int main() { datatype A(12), B(16.44F), C('V'); datatype D = A; A.print(); B.print(); C.print(); D.print(); return 0; }
2.用窮舉法找出1-100之間的所有質數
#include"stdafx.h" #include<iostream> using namespace std; int main() { bool bPrime=true; int iTotalNum=100; for (int i = 1; i <= iTotalNum; i++) { if (i == 1) bPrime = false; for (int j = 2; j<=sqrt(i); j++) { if (i%j == 0) { bPrime = false; break; } } if (bPrime) { cout << i << " "; } bPrime = true; } return 0; }
判斷是否是質數
#include"stdafx.h" #include<iostream> using namespace std; bool Prime(int iNum) { if (iNum == 1) return false; for (int i = 2; i < sqrt(iNum); i++) { if (iNum%i == 0) return false; } return true; } int main() { bool bPrime = true; int iNum; while (cin >> iNum) { if (Prime(iNum)) cout << iNum << "是質數" << endl; else cout << iNum << "不是質數" << endl; } return 0; }
3.編寫函式求兩個函式的最小公倍數和最大公約數
#include"stdafx.h"
#include<iostream>
using namespace std;
int GreatsetCommon(int A, int B);
int LeastCommon(int A, int B,int ileastcommon);
int main()
{
int InterA,InterB;
while (cin >> InterA&&cin >> InterB)
{
if (InterA==0||InterB==0)
{
cout << "最小公約數無，最大公約數為0！" << endl;
continue;
}
int ileastcommon;
ileastcommon= GreatsetCommon(InterA, InterB);
LeastCommon(InterA, InterB,ileastcommon);
}
return 0;
}
int GreatsetCommon(int A,int B)
{
int C;
C = A%B;
while (B!=0)
{
if (C == 0)
{
cout << "最大公約數是：" << B << endl; return B;
}
else
{
A = B; B = C; C = A%B;
}
}
}
int LeastCommon (int A, int B,int ileastcommon)
{
int greastcommon;
if (ileastcommon==0)
{
cout << "無最大公倍數！" << endl;
}
else
{
greastcommon = A*B / ileastcommon;
cout << "最小公倍數為：" << greastcommon <<endl;
return greastcommon;
}
}
4.定義一個矩形類，有長，有寬兩個屬性，有成員函式計算矩形的面積
#include"stdafx.h"
#include<iostream>
using namespace std;
class Rectangle
{
public:
Rectangle(double length,double width);
~Rectangle();
double GetArea();
double GetWidth();
double GetLength();
private:
double dLength;
double dWidth;
};
double Rectangle::GetWidth() {return dWidth;}
double Rectangle::GetArea() {return (dWidth*dLength); }
double Rectangle::GetLength(){return dLength; }
Rectangle::Rectangle(double length, double width)
{
dLength = length; dWidth = width;
}
Rectangle::~Rectangle()
{
}
int main()
{
cout << "請輸入矩形的長和寬：" << endl;
double length, width;
while (cin>>length&&cin>>width)
{
Rectangle rTest= Rectangle(length, width);
cout << "矩形的面積是：" << rTest.GetArea() << endl;
}
return 0;
}
5.定義一個Shape基類，在此基礎上派生出Rectangle和Circle，二者都在GetArea（）函式，計算物件的面積，使用Rectangle類建立一個派生類Square
#include"stdafx.h"
#include<iostream>
using namespace std;
#define PI 3.14
class Shape
{
public:
Shape() {};
~Shape() {};
virtual float Area() = 0;
};
class circle :public Shape
{
public:
circle(float radius) { fRadius = radius; };
~circle() {};
float Area() { return (float)(PI*fRadius*fRadius); };
private:
float fRadius;
};
class rectangle:public Shape
{
public:
rectangle(float length, float width) { fLength = length; fWidth = width; };
~rectangle() {};
float Area() { return fLength*fWidth; }
private:
float fWidth;
float fLength;
};
class square :public rectangle
{
public:
square(float len) :rectangle(len, len) {};
~square() {};
float Area(float len) { return len*len; };
};
int main()
{
Shape *s[3];
s[0] = new circle(2);
cout <<"圓的面積(2)："<< s[0]->Area() << endl;
s[1] = new rectangle(2, 4);
cout << "矩形面積(2,4)："<<s[1]->Area() << endl;
s[2] = new square(3);
cout << "正方形面積：" << s[2]->Area() << endl;
for (int i = 0; i < 3; i++)
{
delete s[i];
}
return 0;
}
6.為了得到一個數的"相反數",我們將這個數的數字順序顛倒,然後再加上原先的數得到"相反數"。例如,為了得到1325的"相反數",首先我們將該數的數字順序顛倒,我們得到5231,之後再加上原先的數,我們得到5231+1325=6556.如果顛倒之後的數字有字首零,字首零將會被忽略。例如n = 100, 顛倒之後是1.
輸入描述:
輸入包括一個整數n,(1 ≤ n ≤ 10^5)
輸出描述:
輸出一個整數,表示n的相反數
輸入例子1:
1325
輸出例子1:
6556
#include<iostream>
#include<vector>
using namespace std;
int main()
{
int iNum;
int iTemp;
int iOpsite=0;
while(cin >> iNum)
{
iTemp = iNum;
while (iTemp)
{
iOpsite = iOpsite * 10 + iTemp % 10;
iTemp /= 10;
}
cout << "相反數" << iOpsite << endl;
cout << iOpsite + iNum;
iNum = 0; iOpsite = 0;
}
return 0;
}