A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example, the name “anna” is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionally numbers can of course be ordered in size. The first few palindrome numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, … The number 10 is not a palindrome (even though you could write it as 010) but a zero as leading digit is not allowed. Input The input consists of a series of lines with each line containing one integer value i (1<= i <= 2*10^9 ). This integer value i indicates the index of the palindrome number that is to be written to the output, where index 1 stands for the first palindrome number (1), index 2 stands for the second palindrome number (2) and so on. The input is terminated by a line containing 0. Output For each line of input (except the last one) exactly one line of output containing a single (decimal) integer value is to be produced. For each input value i the i-th palindrome number is to be written to the output. Sample Input 1 12 24 0 Sample Output 1 33 151
訓練的時候，有一個小步驟推錯了，然後就崩了。 然而這個題其實並不難。 AC代碼：
#include<iostream> #include<cstring> #include<string> #include<cstdio> using namespace std; #define ll long long int int main() { int n; while (~scanf("%d", &n) && n)//n位數 { ll p = 9; int i,t,k ,s[101000];//t=n/2+n%2 t = 1; while (n > p)//9e(t-1) //計算此數在當前位數中的排行~ k的作用是去除前面的0的影響 { t++; n -= p; if (t >= 3 && t % 2 == 1)//從奇數到偶數 任意改變的位數不變！ { p *= 10; } } int len = t / 2 + t % 2;//可以隨意改變的位數 k = p / 9; n += k - 1; for (int i = len; i >= 1; i--)//1/2,1/2的輸出 { s[i] = n % 10; n /= 10; } for (int i = 1; i <= len; i++) cout << s[i]; if (t % 2) len--; for (i = len; i >= 1; i--) cout << s[i]; cout << endl; } return 0; }