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-;
 const int dx[] = {, , -, };
 const int dy[] = {-, , , };
 const ll MOD = 1e9 + ;
 const int MAXN =  + ;
 const int MAXT =  + ;
 using namespace std;
 ll a[];
 void init()
 {
     ll tmp = ll();
     for(int i = ; i < ; i += )
     {
         a[i] = a[i - ] = tmp;
         tmp *= ll();
     }
 }
 ll POW(ll x)
 {
     ll w = ;
     for(ll i = ; i <= x; ++i)
         w *= ll();
     return w;
 }
 int main()
 {
     init();
     int n;
     while(scanf("%d", &n) ==  && n)
     {
         int cnt = ;
         while(n > a[cnt])
         {
             n -= a[cnt];
             ++cnt;
         }
         if(cnt == )
         {
             printf("%d\n", n);
             continue;
         }
         else if(cnt == )
         {
             printf("%d%d\n", n, n);
             continue;
         }
         else
         {
             int tmp = cnt / ;
             char str[];
             memset(str, , sizeof str);
             ll ans = POW(ll(cnt / )) + ll(n - );
             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 %  == ) s.resize(len - );
             reverse(s.begin(), s.end());
             printf("%s\n", s.c_str());
         }
     }
     return ;
 }