http://codeforces.com/contest/725/problem/C
Each English letter occurs at least once in s.
注意到题目有这样一句话,那么就是说S中只有一个重复的字母。一定要看到这句话,不然做不了。
然后就找到pos1和pos2分别代表那两个字母的位置,把他们中间的所有字母对折一下,放在最右边就可以了。因为这样才能用上同一个位置。
只有连续两个相同的字母才会impossible,同样是因为只有两个相同的字母。
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> const int maxn = 26 + 20; char str[111]; int book[maxn]; char ans[maxn][maxn]; void work() { // for (int i = 1; i <= 2; ++i) { // for (int j = 1; j <= 13; ++j) { // ans[i][j] = 'a'; // } // } scanf("%s", str + 1); for (int i = 1; i <= 27; ++i) { book[str[i]]++; } int pos[3]; int cur = 1; for (int i = 1; i <= 27; ++i) { if (book[str[i]] == 2) pos[cur++] = i; } if (pos[1] == pos[2] - 1) { cout << "Impossible" << endl; return; } // cout << pos[1] << " " << pos[2] << endl; int ff = pos[1]; int len = pos[2] - pos[1] - 1; int t = len / 2; ans[1][13 - t] = str[pos[1]]; for (int i = 13 - t + 1; i <= 13; ++i) { ans[1][i] = str[++ff]; } int flag = (len % 2) == 0; for (int i = 13; i >= 13 - t + flag; --i) { ans[2][i] = str[++ff]; } int to = 13 - t - 1; cur = 1; if (to <= 0) { cur = 2; to = 1; } for (int i = pos[2] + 1; i <= 27; ++i) { ans[cur][to] = str[i]; if (cur == 2) to++; else to--; if (to <= 0) { to = 1; cur = 2; } } for (int i = 1; i <= pos[1] - 1; ++i) { ans[cur][to] = str[i]; if (cur == 2) to++; else --to; if (to <= 0) { to = 1; cur = 2; } } for (int i = 1; i <= 2; ++i) { for (int j = 1; j <= 13; ++j) { cout << ans[i][j]; } cout << endl; } } int main() { #ifdef local freopen("data.txt","r",stdin); #endif work(); return 0; }