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  • Project Euler:Problem 55 Lychrel numbers

    If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

    Not all numbers produce palindromes so quickly. For example,

    349 + 943 = 1292,
    1292 + 2921 = 4213
    4213 + 3124 = 7337

    That is, 349 took three iterations to arrive at a palindrome.

    Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

    Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.

    How many Lychrel numbers are there below ten-thousand?

    NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.


    求10000以内的不可按以上方法迭代得出回文的数的个数。

    #include <iostream>
    #include <string>
    using namespace std;
    
    string num2str(int n)
    {
    	string ans = "";
    	while (n)
    	{
    		int a = n % 10;
    		char b = a + '0';
    		ans = b + ans;
    		n /= 10;
    	}
    	return ans;
    }
    
    string strplus(string a, string b)
    {
    	int len = a.length();
    
    	int flag = 0;
    	string ans = "";
    	for (int i = len - 1; i >= 0; i--)
    	{
    		int tmp = a[i] + b[i] - '0' - '0' + flag;
    		flag = tmp / 10;
    		tmp = tmp % 10;
    		char p = tmp + '0';
    		ans = p + ans;
    	}
    	if (flag == 1)
    		ans = '1' + ans;
    	return ans;
    }
    
    bool pali(string a)
    {
    	for (int i = 0; i < a.length() / 2; i++)
    	{
    		if (a[i] != a[a.length() - 1 - i])
    			return false;
    	}
    	return true;
    }
    
    bool isLychrel(int n)
    {
    	string a, b;
    	a = num2str(n);
    	b.assign(a.rbegin(), a.rend());
    	for (int i = 1; i <= 50; i++)
    	{
    		a = strplus(a, b);
    		if (pali(a))
    			return false;
    		b.assign(a.rbegin(), a.rend());
    	}
    	return true;
    }
    
    int main()
    {
    
    	int count = 0;
    	for (int i = 1; i <= 10000; i++)
    	{
    		if (isLychrel(i))
    			count++;
    	}
    	cout << count << endl;
    	system("pause");
    	return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/zfyouxi/p/5280145.html
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