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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 ?rst 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 ?rst 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
题意：给出i，输出第i个回文数，不能有前导0.
题解：第一和最后一位不为0，
我们预处理i位数有多少，以及 在不是第一位和最后一位情况下i位有多少情况
再不断细分就好了
//meek///#include<bits/stdc++.h> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include<iostream> #include<bitset> #include<vector> #include <queue> #include <map> #include <set> #include <stack> using namespace std ; #define mem(a) memset(a,0,sizeof(a)) #define pb push_back #define fi first #define se second #define MP make_pair typedef long long ll; const int N = 1050; const int M = 1000001; const int inf = 0x3f3f3f3f; const int MOD = 1000000007; const double eps = 0.000001; ll a[N],sum[N],b[N],s; void init() {s=0; a[1] = 10;b[1] = 9; a[2] = 10;b[2] = 9; for(int i=3;i<=20;i++) { a[i] = a[i-2] * 10; }s=18; for(int i=3;i<=20;i++) { b[i] = 9*a[i-2]; s+=b[i]; // cout<<s<<endl; } } int main() { init(); ll n, num, ans[N]; while(scanf("%lld",&n)!=EOF) { if(!n) break; mem(ans); int l,r,mm; for(int i=1;i<=20;i++) { if(b[i]>=n) { num = i; mm = num; l = 1, r = num; while(l<=r) { if(num == 1) { if(l!=1) n--; ans[l] = n; ans[r] = n;break; } if(num == 2) { if(l!=1)n--; ans[l] = n; ans[r] = n;break; } ans[l] = n/a[num-2] + (l == 1?1:0); n = n%a[num-2]; if(n == 0 && num-2!=0) { ans[l]--; n = a[num-2]; } ans[r] = ans[l]; l++,r--;num-=2; } break; } n -= b[i]; } for(int i=1;i<=mm;i++) printf("%lld",ans[i]); printf("\n"); } return 0; }
POJ2402/UVA 12050 Palindrome Numbers 数学思维
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原文地址：http://www.cnblogs.com/zxhl/p/5080168.html
