poj 2402 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

这道题得找区间，1位数和2位数的回文数都是9个，3位数和4位数的回文数都是9*10个，5位和6位都是9*10*10个，那么就可以找到所求数字的位数了，然后就是从小到大二分查找所求的数了，并且只要找前一半就好了。

#include <iostream> #include <cstdio> using namespace std; long long int find(long long int n){ long long int ans,i,j,pos,cnt,tp,left,right; cnt=0; tp=9; for(i=1;i;i++){ if(cnt+tp>=n){ pos=i; break; } cnt+=tp; if(i%2!=1) tp*=10; } n-=cnt; left=1; for(i=2;i<=(pos+1)/2;i++){ left*=10; } ans=left+=n-1; if(pos%2==1) left/=10; while(left){ ans=ans*10+left; left/=10; } return ans; } int main(){ long long int n; while(cin>>n,n){ printf("%I64d\n",find(n)); } return 0; }