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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