Project Euler 1-10练习题希望自己坚持下去

附上在线IDE http://www.shucunwang.com/RunCode/cpp/

1  求1000以内所有3或5的倍数的和

#include 
using namespace std;

int main() {
    
    int sum = 0;
    for(int i=0; i<1000; ++i)
    {
        if(i%3==0 || i%5==0)
        {
            sum += i;
        }
    }
    cout << "sum:" << sum << endl;
	return 0;
}

2  斐波那契数列中不超过四百万的项，求其中为偶数的项之和

#include 
using namespace std;


int fibonacci(int first, int second, int& total_sum)
{
    if(first <= 4000000)
    {
        int tmp_sum = first + second;
        total_sum += first%2==0?first:0;
        return fibonacci(second, tmp_sum, total_sum);
    }
    return first;
}


int main() {
    
    int sum = 0;
    fibonacci(1,2,sum);
    cout << "even-sum:" << sum << endl;
	return 0;
}

3   600851475143最大的质因数是多少

#include 
using namespace std;


int largest_prime_factor(long n)
{
  int min = 2;
  while (n > min) {
    for (int i = min; i <= n; i++) {
      if (n % i == 0) {
        n = n / i;
        min = i;
        break;
      }
    }
  }
  return min;
}


int main() {
  int max_prime_factor = largest_prime_factor(600851475143);
  cout  << "max:" << max_prime_factor << endl;
  return 0;
}

4 由两个3位数相乘得到的最大回文乘积

#include 
#include 
using namespace std;

bool is_palindromic(int num)
{
    std::ostringstream out;
    out << num;
    string num_str = out.str();
    int max_size = num_str.size();
    for(int i=0; i=1; --i)
    {
        for(int j=999; j>=1; --j)
        {
            int mul_nm = i*j;
            bool is_palindromic_num = is_palindromic(mul_nm);
            if(is_palindromic_num)
            {
                max = mul_nm > max ? mul_nm : max;
            }
        }
    }
    return max;
}


int main()
{
    int max_palindromic = get_max_palindromic();
    cout << "result:" << max_palindromic << endl;
    return 0;
}

5. 最小的能够被1到20整除的正数--最小倍数

#include 
#include 

6.平方的和与和的平方之差

#include 
#include 
#include 
using namespace std;

int main()
{
	int square_sum = 0;
    int sum_square = 0;
    for(int i=1; i<=100; ++i)
    {
        square_sum += pow(i*1.0,2);
        sum_square += i;
    }
    sum_square = pow(sum_square*1.0,2);
    int diff_val = abs(square_sum - sum_square);
    cout << "diff value: " << diff_val << endl;
	return 0;
}

7 第10001个质数

#include 
#include 
using namespace std;

int get_prime(int n)
{
	int *p_r = new int[n];
	int count = 1;
	p_r[0] = 2;
	p_r[1] = 3;

	for(int i=3; count!=n; i+=2)
	{
		int sqrt_i = floor(sqrt(i*1.0));
        int j = 1;
		for(j=1; sqrt_i>=p_r[j] && i%p_r[j]!=0; ++j);

		if(sqrt_i

8.连续数字最大乘积

#include 
#include 
#include 
using namespace std;

//找出连续n位最大的字符串
long get_max_num(const string& src, int n)
{
    if (src.empty())
    {
       return 0;
    }
    string max_str = src.substr(0, n);
    for (int idx=1; idx

9.特殊毕达哥拉斯三元组

#include 
#include 
using namespace std;

int main() {
    int num1,num2,num3;
    for(int i=1; i<1000; ++i)
    {
        for(int j=1;j<1000; ++j)
        {
            int tmp = i*i + j*j;
            int k = sqrt(1.0*tmp);
            if(i+j+k == 1000 && tmp==k*k && i

10.素数的和

#include 
#include 
using namespace std;

bool is_prime(int num)
{
    if (num < 2)
    {
        return false;
    }
    int tmp = sqrt(1.0*num);
    for (int i=2; i<=tmp; ++i)
    {
        if (num%i == 0)
        {
            return false;
        }
    }
    return true;
}

int main() {

    long long sum = 0;
    for (int i=2; i<=2000000; ++i)
    {
        if (is_prime(i))
        {
            sum += i;
        }
    }
    cout << "sum:" << sum << endl;
    return 0;
}