C++課後題(1)
阿新 • • 發佈：2018-12-13
使用級數計算圓周率
#include <iostream> #include <cmath> #include <iomanip> using namespace std; float m(int); int main(int argc, char *argv[]) { cout << "i" << " " << "m(i)" << endl; //cout << m(10) << endl; for(int j=1; j<10000; j++) { cout <<setw(5) << j << " " << setw(15) << m(j) << endl; } return 0; } float m(int n) { float sum = 0; for(int i=1; i<=n; i++) { sum += pow(-1.0, i+1)*1.0/(2*i-1); } return 4.0*sum; }
尋找孿生素數：
素數a, 素數b, 若b-a=2, 則a,b為孿生素數
#include <iostream> #include <cmath> #include <iomanip> using namespace std; bool isPrime(int); //bool isTwinPrime(int); int main(int argc, char *argv[]) { for(int j=2; j<=1000; j++) { if(isPrime(j) && isPrime(j+2)) { cout << "(" << j << " , " << j+2 << ")" << endl; } } return 0; } bool isPrime(int n) { bool prime = true; for(int i=2; i<n; i++) { if(n%i==0) { prime = false; break; } } return prime; }
尋找回文素數：
判斷一個數是否為迴文數，有多種方法，常見的有三種：
1.將輸入數字轉化為字串。迴文數關於中心對稱，只需比較對稱的數是否相等即可。
2.採用除10取餘的方法，從最低位開始，依次取出該數的各位數字，然後用最低為充當最高位，按反序構成新的數，再與原數比較。
3.採用棧的方式。判斷出棧的元素與棧內字元是否相等。
#include <iostream> #include <cmath> #include <iomanip> using namespace std; bool isPrime(int); bool isPalindromicPrime(int); bool isPalindromic(int n); //bool isTwinPrime(int); int main(int argc, char *argv[]) { //cout << isPalindromic(100097) << endl; int count = 0; int k = 2; while(count<100) { if(isPalindromicPrime(k)) { count++; cout << setw(5) << k << endl; //if(count%10==0) cout << endl; } k++; } return 0; } bool isPrime(int n) { bool prime = true; for(int i=2; i<n; i++) { if(n%i==0) { prime = false; break; } } return prime; } bool isPalindromic(int n) { unsigned int i=n; unsigned int m=0; while(i>0) { m = m*10 + i%10; i = i / 10; } return m==n; } bool isPalindromicPrime(int n) { bool PalindromicPrime = false; if(isPrime(n)) { if(isPalindromic(n)) { PalindromicPrime = true; } } return PalindromicPrime; }
6. 巴比倫方法實現sqrt()函式
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdlib>
#include <time.h>
using namespace std;
double cal_sqrt(int );
int main(int argc, char *argv[])
{
cout << "Enter a number: ";
int nn;
cin >> nn;
double result = cal_sqrt(nn);
cout << "sqrt(" << nn << ")" << " is " << result << endl;
return 0;
}
double cal_sqrt(int n) // 巴比倫方法實現開跟運算
{
srand(time(0));
//double last_guess = static_cast<double>(rand() % 10); // generate random number[0, 10)
double last_guess = 1.0;
double next_guess = (last_guess + (n / last_guess)) / 2.0;
int count = 0;
cout << "Iteration" << " " << "sqrt" << endl;
while(abs(last_guess - next_guess) > 0.000001)
{
last_guess = next_guess;
next_guess = (last_guess + (n / last_guess)) / 2.0;
count++;
cout << count << " " << next_guess << endl;
}
return next_guess;
}
利用級數來逼近e
e = 1 + 1/1! + 1/2! + 1/3! + .......+ 1/i!
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int jiecheng(int);
double calculate_e(int);
int main(int argc, char *argv[])
{
cout << calculate_e(20) << endl;
return 0;
}
double calculate_e(int n)
{
double sum = 0.0;
for(int k=0; k<=n; k++)
{
sum += 1.0 / jiecheng(k);
}
return sum;
}
int jiecheng(int n)
{
int a = 1;
if(n==0)
{
return a;
}
else
{
for(int i=n; i>0; i--)
{
a = a*i;
}
return a;
}
}
實現簡單的計算器：
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdlib>
#include <time.h>
using namespace std;
void display_menu();
int main(int argc, char *argv[])
{
while(true)
{
srand(time(0));
display_menu();
int answer;
cout << "Please Enter Your Choice: ";
cin >> answer;
float x1, x2;
x1 = rand() % 10; // Generate the first random number
x2 = rand() % 10; // The second
if(answer==1)
{
cout << "What is " << x1 << " + " << x2 << " ? " << endl;
cout << "Your answer: ";
float tmp;
cin >> tmp;
if(tmp==(x1+x2))
{
cout << "You are correct!" << endl;
}
else
{
cout << "YUou are wrong!" << endl;
}
}
if(answer==2)
{
cout << "What is " << x1 << " - " << x2 << " ? " << endl;
cout << "Your answer: ";
float tmp;
cin >> tmp;
if(tmp==(x1-x2))
{
cout << "You are correct!" << endl;
}
else
{
cout << "YUou are wrong!" << endl;
}
}
if(answer==3)
{
cout << "What is " << x1 << " x " << x2 << " ? " << endl;
cout << "Your answer: ";
float tmp;
cin >> tmp;
if(tmp==(x1*x2))
{
cout << "You are correct!" << endl;
}
else
{
cout << "YUou are wrong!" << endl;
}
}
if(answer==4)
{
while(x2==0)
{
x2 = rand() % 10;
}
cout << "What is " << x1 << " / " << x2 << " ? " << endl;
cout << "Your answer: ";
float tmp;
cin >> tmp;
if(tmp==(x1/x2))
{
cout << "You are correct!" << endl;
}
else
{
cout << "YUou are wrong!" << endl;
}
}
if(answer==5)
{
exit(0);
}
}
return 0;
}
void display_menu()
{
cout << " Main menu " << endl;
cout << " 1. Addition " << endl;
cout << " 2. Substraction " << endl;
cout << " 3. Multiply " << endl;
cout << " 4. Division " << endl;
cout << " 5. Exit " << endl;
}
密碼有效性檢驗： 假定密碼的規則為：
1.密碼至少有八個字元
2.密碼必須僅包含字母和數字
3.密碼必須包含至少兩個數字
code:
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdlib>
#include <time.h>
#include <string>
#include <cctype> // 包含字元處理函式，大小寫轉換 ， 大小寫轉換的方法
using namespace std;
int main(int argc, char *argv[])
{
cout << "Enter you password: ";
string password;
cin >> password;
if(password.size()<8)
{
cout << "password " << password << " is not valid" << endl;
cout << "length " << password.size() << endl;
}
else // 滿足條件1
{
int flag=0;
int count_number = 0;
for(int index=0; index<password.length(); index++)
{
if((tolower(password[index])>='0' && tolower(password[index])<='9') || (tolower(password[index])>='a' || tolower(password[index])<='z'))
{
if(password[index]>='0' && password[index]<='9')
count_number++; // 統計數字的數量
}
else
{
flag = 1;
break;
}
}
// cout << "flag " << flag << "count_number " << count_number << endl;
if(flag==0 && count_number>=2)
{
cout << "password " << password << " is valid" <<endl;
}
else
{
cout << "password " << password << " is not valid" <<endl;
}
}
return 0;
}
問題：
求解一元二次方程：
ax*x+bx+c=0
根據求根公式編寫程式碼：
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdlib>
#include <time.h>
#include <string>
#include <cctype> // 包含字元處理函式，大小寫轉換 ， 大小寫轉換的方法
using namespace std;
int main(int argc, char *argv[])
{
cout << "Solve the equation: ax*x+bx+c=0" << endl;
cout << "Enter the coeffient of the equation(a!=0): ";
int a, b, c;
cin >> a >> b >> c;
int delta = pow(b,2.0) - 4*a*c;
if(delta>=0) // 方程有實數根
{
if(delta>0) // 兩個不同的實根
{
float r1, r2;
r1 = (-b+sqrt(delta))/2*a;
r2 = (-b-sqrt(delta))/2*a;
cout << "r1=" << r1 << endl;
cout << "r2=" << r2 << endl;
}
else // 兩個相等的實數根
{
float r;
r = (-b)/(2*a);
cout << "The solution of the equation is: " << endl;
cout << "r1=" << r << endl;
cout << "r2=" << r << endl;
}
}
else // 虛數根
{
delta = -1 * delta;
float real, imaginary;
real = -1.0*b/(2*a);
imaginary = sqrt(delta) / (2.0*a);
cout << "The solution of the equation is: " << endl;
cout << "r1=" << real << "+" << imaginary << "i" << endl;
cout << "r2=" << real << "-" << imaginary << "i" << endl;
}
return 0;
}
測試結果：
判斷兩個直線是否相交，如果相交，求解兩個直線的交點
#include <iostream>
#include <ctime>
#include <cmath>
using namespace std;
// 判斷兩條直線的交點
void cal_Intersection(double, double, double, double, double, double, double, double, double&, double&, bool&);
int main(int argc, char *argv[])
{
double x, y; // intersection point
bool intersection;
cout << "輸入直線的端點座標： " << endl;
cout << "Line 1: ";
double l1_x1, l1_y1, l1_x2, l1_y2;
cin >> l1_x1 >> l1_y1 >> l1_x2 >> l1_y2;
cout << "Line 2: ";
double l2_x1, l2_y1, l2_x2, l2_y2;
cin >> l2_x1 >> l2_y1 >> l2_x2 >> l2_y2;
cal_Intersection(l1_x1, l1_y1, l1_x2, l1_y2, l2_x1, l2_y1, l2_x2, l2_y2, x, y, intersection);
cout << intersection << endl;
if(intersection)
{
cout << "The cross point is (" << x << "," << y << ")." << endl;
//cout << x << " " << y;
}
else
{
cout << "The two line has no cross point!" << endl;
//cout << x << " " << y;
}
return 0;
}
void cal_Intersection(double x1, double y1, double x2, double y2, double x3, double y3, double x4, double y4, double& x, double& y, bool& isIntersection)
{
// 根據直線的方程
// ax+by=e
// cx+dy=f
double a = y2 - y1;
double b = -1 * (x2 - x1);
double e = x2*(y2 - y1) - y2*(x2 - x1);
double c = y4 - y3;
double d = -1 * (x4 - x3);
double f = x4*(y4 - y3) - y4*(x4 - x3);
if(abs(a*d - b*c)<0.0001) // 方程無解
{
x = 0;
y = 0;
isIntersection = false;
}
else // 方程有解
{
x = (e*d - b*f) / (a*d - b*c);
y = (a*f - e*c) / (a*d - b*c);
if(x1 < x2) // 消除點的先後次序對結果的影響
{
if(x>x1 && x<x2)
isIntersection = true;
else
isIntersection = false;
}
else
{
if(x>x2 && x<x1)
isIntersection = true;
else
isIntersection = false;
}
}
}
信用卡號碼校驗： 開頭4： visa card
開頭5： mastercard card
開頭37： America express card
開頭6： Discover card;
Luhn校驗法：
判斷信用卡號是否合法：
code:
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;
int cardTypeCode(string);
void cardType(string, string&);
int sumOfEven(string);
int sumOfOdd(string);
bool isValid(string);
int main(int argc, char *argv[])
{
// 信用卡號碼校驗
string card_number;
string card_type;
cout << "Enter your card number: ";
cin >> card_number;
if(isValid(card_number))
{
cardType(card_number, card_type);
cout << "The card is valid, the card type is " << card_type << endl;
}
else
{
cout << "The card is invalid" << endl;
}
return 0;
}
int cardTypeCode(string card)
{
// 根據信用卡的卡號返回卡的型別
if(card[0]=='4') // Visa card
return 1;
else if(card[0]=='5') // Mastercard
return 2;
else if(card[0]=='3' && card[1]=='7') // America express card
return 3;
else if(card[0]=='6') // Discovery card
return 4;
else // other card
return 0;
}
void cardType(string card, string& cardTypeString)
{
switch (cardTypeCode(card))
{
case 0: cardTypeString = "Other card"; break;
case 1: cardTypeString = "Visa card"; break;
case 2: cardTypeString = "MasterCard card"; break;
case 3: cardTypeString = "America express card"; break;
case 4: cardTypeString = "Discover card"; break;
}
}
int sumOfEven(string card)
{
// 從右向左偶位數*2
int sum_even = 0;
int counter = 0; // 計數器
for(int i=card.length()-1; i>=0; i--)
{
int current_number = 0 + card[i] - '0';
counter++;
if(counter%2==0) // 偶數位的數字
{
//cout << "current number: " << current_number << endl;
if(2*current_number>9)
{
current_number = current_number * 2;
int a = current_number % 10; // 儲存各位
int b = current_number / 10; // 儲存十位數
sum_even += (a+b);
}
else
{
sum_even += current_number*2;
}
}
}
return sum_even;
}
int sumOfOdd(string card)
{
int sum_odd = 0;
int counter = 0;
for(int i=card.length()-1; i>=0; i--)
{
int current_number = 0 + card[i] - '0';
counter++;
if(counter%2==1)
{
sum_odd += current_number;
}
}
return sum_odd;
}
bool isValid(string card)
{
if((sumOfEven(card)+sumOfOdd(card))%10==0)
return true;
else
return false;
}