C++课后题(1)
使用级数计算圆周率
#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;
}
寻找孪生素数：
素数a, 素数b, 若b-a=2, 则a,b为孪生素数
#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;
}
寻找回文素数：
判断一个数是否为回文数，有多种方法，常见的有三种：
1.将输入数字转化为字符串。回文数关于中心对称，只需比较对称的数是否相等即可。
2.采用除10取余的方法，从最低位开始，依次取出该数的各位数字，然后用最低为充当最高位，按反序构成新的数，再与原数比较。
3.采用栈的方式。判断出栈的元素与栈内字符是否相等。
#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. 巴比伦方法实现sqrt()函数
#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) // 巴比伦方法实现开跟运算
{
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;
}
利用级数来逼近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;
}
}
实现简单的计算器：
#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;
}
密码有效性检验：
假定密码的规则为：
1.密码至少有八个字符
2.密码必须仅包含字母和数字
3.密码必须包含至少两个数字
code:
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdlib>
#include <time.h>
#include <string>
#include <cctype> // 包含字符处理函数，大小写转换 ， 大小写转换的方法
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 // 满足条件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++; // 统计数字的数量
}
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;
}
问题：
求解一元二次方程：
ax*x+bx+c=0
根据求根公式编写代码：
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdlib>
#include <time.h>
#include <string>
#include <cctype> // 包含字符处理函数，大小写转换 ， 大小写转换的方法
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) // 方程有实数根
{
if(delta>0) // 两个不同的实根
{
float r1, r2;
r1 = (-b+sqrt(delta))/2*a;
r2 = (-b-sqrt(delta))/2*a;
cout << "r1=" << r1 << endl;
cout << "r2=" << r2 << endl;
}
else // 两个相等的实数根
{
float r;
r = (-b)/(2*a);
cout << "The solution of the equation is: " << endl;
cout << "r1=" << r << endl;
cout << "r2=" << r << endl;
}
}
else // 虚数根
{
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;
}
测试结果：
判断两个直线是否相交，如果相交，求解两个直线的交点
#include <iostream>
#include <ctime>
#include <cmath>
using namespace std;
// 判断两条直线的交点
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 << "输入直线的端点坐标： " << 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)
{
// 根据直线的方程
// 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) // 方程无解
{
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) // 消除点的先后次序对结果的影响
{
if(x>x1 && x<x2)
isIntersection = true;
else
isIntersection = false;
}
else
{
if(x>x2 && x<x1)
isIntersection = true;
else
isIntersection = false;
}
}
}
信用卡号码校验：
开头4： visa card
开头5： mastercard card
开头37： America express card
开头6： Discover card;
Luhn校验法：
判断信用卡号是否合法：
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[])
{
// 信用卡号码校验
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)
{
// 根据信用卡的卡号返回卡的类型
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)
{
// 从右向左偶位数*2
int sum_even = 0;
int counter = 0; // 计数器
for(int i=card.length()-1; i>=0; i--)
{
int current_number = 0 + card[i] - '0';
counter++;
if(counter%2==0) // 偶数位的数字
{
//cout << "current number: " << current_number << endl;
if(2*current_number>9)
{
current_number = current_number * 2;
int a = current_number % 10; // 存储各位
int b = current_number / 10; // 存储十位数
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;
}
