POJ - 2042 - Palindrome Numbers

Palindrome Numbers Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3948 Accepted: 1505

Description

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

Source

Tehran 2003 Preliminary

题意：求第n个回文串数字（1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,...），其中不存在有前导零的数

n位数和 n+1（n为奇数） / n-1（n为偶数）位数所含有的回文串数字个数相同，均为(10^(n/2+n%2-1))*9，因为除最外层只有1-9九种选择，其他位都是有0-9十种选择。

所以我们可以先确定位数，再确定各个位置上的数字。

因为是对称的，所以我们只用求一半的数字就行了，设一半的位数位haf，第 i 位上的数字 +1 就是遍历过了10^(i-1)种数字，因为是按从小到大的顺序来的，所以先变的是中间的数，然后再往两边推，所以我们就反过来遍历就行了。

#include<iostream> #include<stdio.h> #include<math.h> #include<string.h> using namespace std; #define LL long long LL n,a[15]; int main() { a[0] = 1; for(int i=1; i<=10; i++) a[i] = pow(10,i-1)*9; while(cin>>n,n) { int i; for(i=1;;i++) { if(n-a[i/2+i%2]<=0) break; n-=a[i/2+i%2]; } int haf = i/2+i%2,num[1111];//我取的是后半部分 int p = i%2; memset(num,0,sizeof num); num[haf] = 1;//最后一位只能从1开始 n--; while(n) { for(int i=haf; i>=1; i--) { if(n>=pow(10,i-1)) num[i]+=n/pow(10,i-1); n%=(LL)pow(10,i-1); } } for(int i=haf; i>=1; i--) cout<<num[i]; for(int i=1+p;i<=haf;i++) cout<<num[i]; cout<<endl; } return 0; }