UVA 12050 Palindrome Numbers

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

思路  1，2，3，4，5，6，7，8，9

          11，22，33，44，55，66，77，88，99

xox                9*10;中间可以填零

xoox              9*10;

xooox            9*10*10

xoooox          9*10*10

xooooox         9*10*10*10

xoooooox       9*10*10*10

规律已经出来了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
ll a[10],b[20],cnt[20];
void init()
{
    a[0]=1;
    for(int i=1;i<10;i++){
        a[i]=a[i-1]*10;
    }
    for(int i=0;i<20;i+=2){
        b[i]=b[i+1]=a[i/2]*9;
    }
    cnt[0]=9;
    for(int i=1;i<20;i++){
        cnt[i]=cnt[i-1]+b[i];
    }
}
int main()
{
    ll n,pos,ans,inf;
    init();
    while(cin>>n && n)
    {
        pos=0;
        int s[35];
        pos=lower_bound(cnt,cnt+20,n)-cnt;
        ans=a[pos/2]+(pos>0?n-1-cnt[pos-1]:n-1);
        inf=0;
        while(ans)
        {
            s[inf++]=ans%10;
            ans/=10;
        }
        for(int i=inf-1;i>=0;i--)
            printf("%d",s[i]);
        for(int i=pos%2?0:1;i<inf;i++)
            printf("%d",s[i]);
        printf("\n");
    }
    return 0;
}

 再来一个反向求？

"回文数"是一种数 字。如：98789, 这个数字正读是98789,倒读也是98789,正读倒读一样，所以这个数字就是回文数。现在给你一个回文数n(0<n<10¹⁹)，让你求出这个数是第几个的回文正整数。

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
ll a[10],b[20],num[20];
void ac()
{
    a[0]=1;
    for(int i=1;i<10;i++){
        a[i]=a[i-1]*10;
    }
    for(int i=0;i<20;i+=2){
        b[i]=b[i+1]=a[i/2]*9;
    }
    num[0]=9;
    for(int i=1;i<20;i++){
        num[i]=num[i-1]+b[i];
    }
}
int find(int n)
{
    ac();
    int pos[20];
    int cnt=0,k;
    while(n)
    {
        pos[cnt++]=n%10;
        n/=10;
    }
    if(cnt%2==1) k=cnt/2;
    else k=cnt/2-1;
    int ans=0;
    int m=cnt-1;
    for(int i=0;i<=k;i++)
    {
        ans=ans*10+pos[i];
    }
    if(m==0)return ans-a[m/2]+1;
    else return ans-a[m/2]+num[m-1]+1;
}
int main()
{
    int n;
    while(cin>>n)
    {
        cout<<find(n)<<endl;
    }
    return 0;
}