UVALive 2889（回文数字）

题意：给定i，输出第i个回文数字。

分析：1,2,3,4，……，9------------------------------------------------------------------------------------------9个

　　　11,12,13,14，……，19-------------------------------------------------------------------------------------9个

　　　101,111,121,131,141,151,161,171,181,191,202,212,222,232，……，979,989,999-------------------------------90个

　　　1001,1111,1221,1331,1441，……，9889,9999----------------------------------------------------------------90个

　　　10001,10101,10201,10301，……，99899,99999--------------------------------------------------------------900个

　　　规律是把回文串一分为二看，例如第四行，前两个数字是从10到99，共90个数字。而第三行也是从10到99，区别在于，需要去掉最后一位再反转，才是另一半（后两个数字）。

思路：先确定i所对应的回文数字的位数，再确定具体值。

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << "   "
#define prln(x) cout << #x << " : " << x << endl
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {-1, 1, 0, 0};
const ll MOD = 1e9 + 7;
const int MAXN = 1000000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
ll a[25];
void init()
{
    ll tmp = ll(9);
    for(int i = 1; i < 20; i += 2)
    {
        a[i] = a[i - 1] = tmp;
        tmp *= ll(10);
    }
}
ll POW(ll x)
{
    ll w = 1;
    for(ll i = 1; i <= x; ++i)
        w *= ll(10);
    return w;
}
int main()
{
    init();
    int n;
    while(scanf("%d", &n) == 1 && n)
    {
        int cnt = 0;
        while(n > a[cnt])
        {
            n -= a[cnt];
            ++cnt;
        }
        if(cnt == 0)
        {
            printf("%d\n", n);
            continue;
        }
        else if(cnt == 1)
        {
            printf("%d%d\n", n, n);
            continue;
        }
        else
        {
            int tmp = cnt / 2;
            char str[50];
            memset(str, 0, sizeof str);
            ll ans = POW(ll(cnt / 2)) + ll(n - 1);
            sprintf(str, "%lld", ans);//把数字变为指定格式的字符串
            printf("%s", str);
            string s = string(str);
            int len = s.size();
            if(cnt % 2 == 0) s.resize(len - 1);
            reverse(s.begin(), s.end());
            printf("%s\n", s.c_str());
        }
    }
    return 0;
}