• Project Euler Problem (1~10)


    1.If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.Find the sum of all the multiples of 3 or 5 below 100

    (计算1000以内能被3或5整除的数之和)

    #include <iostream>
    using namespace std;
    int main()
    {
    	int i;
    	int sum = 0;
    	for (i = 1; i <= 1000; i++)
    	{
    		if (i % 3 == 0 || i % 5 == 0)
    			sum = sum + i;
    	}
    	cout << sum << endl;
    	return 0;
    }
    
    运行结果:
    234168
    

    2.Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...Find the sum of all the even-valued terms in the sequence which do not exceed one million.

    (计算斐波那契数列中小于100万的偶数项之和)

    #include <iostream>
    using namespace std;
    int main()
    {
    	int a = 1, b = 2;
    	int sum = 0;
    	while(a < 1000000 && b < 1000000)
    	{
    		if (a % 2 == 0)
    			sum = sum + a;
    		if (b % 2 == 0)
    			sum = sum + b;
    		a = a + b;
    		b = b + a;
    	}
    	cout << sum << endl;
    	return 0;
    }
    
    运行结果:
    257114
    

    3.The prime factors of 13195 are 5, 7, 13 and 29.What is the largest prime factor of the number 317584931803?

    (分解出317584931803的因数)

    #include <iostream>
    using namespace std;
    int main()
    {
    	int i;
    	long long n = 317584931803;
    	for (i = 2; i <= n; i++)
    	{
    		while (n != i)
    		{
    			if (n % i == 0)
    			{
    				n = n / i;
    			}
    			else
    				break;
    		}
    	} //从2开始将每个数作为n的除数,若整除则为因数后重新计算n再判断
    	cout << n << endl;//前面只输入前面几个较小因数,重新计算的最后一个n为最大因数
    	return 0;
    }
    
    运行结果:
    3919
    

    4.A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.Find the largest palindrome made from the product of two 3-digit numbers.

    (找出两个三位数相乘的最大回文数)

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  • 原文地址:https://www.cnblogs.com/kkyblog/p/11189541.html
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