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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。
1 #include<cstdio> 2 #include<cstring> 3 #include<cctype> 4 #include<cstdlib> 5 #include<cmath> 6 #include<iostream> 7 #include<sstream> 8 #include<iterator> 9 #include<algorithm> 10 #include<string> 11 #include<vector> 12 #include<set> 13 #include<map> 14 #include<deque> 15 #include<queue> 16 #include<stack> 17 #include<list> 18 #define fin freopen("in.txt", "r", stdin) 19 #define fout freopen("out.txt", "w", stdout) 20 #define pr(x) cout << #x << " : " << x << " " 21 #define prln(x) cout << #x << " : " << x << endl 22 typedef long long ll; 23 typedef unsigned long long llu; 24 const int INT_INF = 0x3f3f3f3f; 25 const int INT_M_INF = 0x7f7f7f7f; 26 const ll LL_INF = 0x3f3f3f3f3f3f3f3f; 27 const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; 28 const double pi = acos(-1.0); 29 const double EPS = 1e-6; 30 const int dx[] = {0, 0, -1, 1}; 31 const int dy[] = {-1, 1, 0, 0}; 32 const ll MOD = 1e9 + 7; 33 const int MAXN = 1000000 + 10; 34 const int MAXT = 10000 + 10; 35 using namespace std; 36 ll a[25]; 37 void init() 38 { 39 ll tmp = ll(9); 40 for(int i = 1; i < 20; i += 2) 41 { 42 a[i] = a[i - 1] = tmp; 43 tmp *= ll(10); 44 } 45 } 46 ll POW(ll x) 47 { 48 ll w = 1; 49 for(ll i = 1; i <= x; ++i) 50 w *= ll(10); 51 return w; 52 } 53 int main() 54 { 55 init(); 56 int n; 57 while(scanf("%d", &n) == 1 && n) 58 { 59 int cnt = 0; 60 while(n > a[cnt]) 61 { 62 n -= a[cnt]; 63 ++cnt; 64 } 65 if(cnt == 0) 66 { 67 printf("%d\n", n); 68 continue; 69 } 70 else if(cnt == 1) 71 { 72 printf("%d%d\n", n, n); 73 continue; 74 } 75 else 76 { 77 int tmp = cnt / 2; 78 char str[50]; 79 memset(str, 0, sizeof str); 80 ll ans = POW(ll(cnt / 2)) + ll(n - 1); 81 sprintf(str, "%lld", ans);//把数字变为指定格式的字符串 82 printf("%s", str); 83 string s = string(str); 84 int len = s.size(); 85 if(cnt % 2 == 0) s.resize(len - 1); 86 reverse(s.begin(), s.end()); 87 printf("%s\n", s.c_str()); 88 } 89 } 90 return 0; 91 }
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原文地址：http://www.cnblogs.com/tyty-Somnuspoppy/p/5906466.html
