POJ2402/UVA 12050 Palindrome Numbers 数学思维

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;
}

代码