题目详情
如果一个数正着读和反着读一样大,则这个数叫做回文数,例如121是回文数,123454321是回文数。
现给定一个正整数x,输出一个回文数y,要求y > x,并且组成x的所有数字之和与组成y的所有数字之和相等,以及y > x。
x在10^1000以内,因为数字较大,我们用字符串作为输入和输出。
如果无解,请输出Impossible。如果有多个y,输出最小的那个。
例如:
输入919,输出14941
输入1,输出Impossible
class Program { static void Main(string[] args) { string str = "919"; int sum = Sum(str); if (str == "1" || str == "0" || str[0] == '-') { Console.WriteLine("Impossible"); } do { str = Next(str); if (IsHuiWen(str) && Sum(str) == sum) { Console.Write(str); break; } } while (true); Console.Read(); } /// <summary> /// Sum of each digit in a string. /// </summary> /// <param name="str">The string.</param> /// <returns></returns> static int Sum(string str) { int s = 0; int low = 0; while (low < str.Length) { s += str[low++] - '0'; } return s; } /// <summary> /// whether a string is huiwen. /// </summary> /// <param name="str">The string.</param> /// <returns></returns> static bool IsHuiWen(string str) { int low = 0; int high = str.Length - 1; while (low < high) { if (str[low++] != str[high--]) return false; } return true; } /// <summary> /// String plus 1. /// </summary> /// <param name="str"></param> /// <returns></returns> static string Next(string str) { char[] cs = new char[str.Length + 1]; int i = str.Length - 1; int j = 0; int a = 1; int t = 0; while (i >= 0) { t = str[i--] - '0' + a; cs[j++] = (t % 10).ToString()[0]; a = t / 10; } if (a == 1) { cs[j] = '1'; } string s = ""; int high = cs.Length - 1; while (high >= 0) { if (cs[high] == '