Dhaka 2003 / UVa 12050 - Palindrome Numbers (回文數)

12050 - Palindrome Numbers

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=467&page=show_problem&problem=3202

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*109) . 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      

打表，先根據表找出這個回文數有多少位，然後算一般，另一半照前面的輸出。

完整代碼：

/*0.015s*/

#include<cstdio>
#include<cstring>
#include<cstdlib>
const int endsum[21] =
{
	0, 9, 18, 108, 198, 1098, 1998, 10998, 19998, 109998, 199998, 1099998, 1999998,
	10999998, 19999998, 109999998, 199999998, 1099999998, 1999999998, 2000000001
};

char str[20];

int main(void)
{
	int len, num, i, j, maxj;
	int ans;
	while (gets(str), str[0] & 15)
	{
		len = strlen(str);
		num = atoi(str);
		if (len == 1)
			putchar(str[0]);
		else if (num < 19)
		{
			num += 39;
			putchar(num);putchar(num);
		}
		else
		{
			i = 2;
			while (endsum[++i] < num)
				;
			/// 此時回文數的位數為i
			num -= endsum[i - 1] + 1; /// 多減一個
			/// 現在num是第num個i位數的回文數
			ans = 1;
			maxj = (i + 1) >> 1; /// i/2上取整
			for (j = 1; j < maxj; ++j)
				ans *= 10;
			ans += num;
			sprintf(str, "%d", ans);
			for (j = 0; j < maxj; ++j)
				putchar(str[j]);
			if ((i & 1) == 0)///優先級啊。。
				putchar(str[maxj - 1]);
			for (j = maxj - 2; j >= 0; --j)
				putchar(str[j]);
		}
		putchar('\n');
	}
	return 0;
}