题意：一个字符串被称为square当且仅当它可以由两个相同的串连接而成. 例如, "abab", "aa"是square, 而"aaa", "abba"不是. 两个长度相同字符串之间的
hamming distance是对应位置上字符不同的位数. 给定一行字符串和 m，输出字典序最小的字符串。
析：首先先用dp判断能不能形成这样的字符串，然后再打印出来，dp[i][j] 表示 i - 中间的数能不能改 j 个字符得到，最后打印时贪心，从最小的开始，如果能用结束。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn/2][maxn];
char s[maxn], ans[maxn]; int main(){
int T; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
scanf("%s", s+1);
memset(dp, false, sizeof dp); int mid = n/2;
dp[mid+1][0] = true;
for(int i = mid; i >= 1; --i){
if(s[i] == s[n-mid+i]){
for(int j = 0; j <= m; ++j){
if(j > 1) dp[i][j] |= dp[i+1][j-2];
dp[i][j] |= dp[i+1][j];
}
}
else{
for(int j = 0; j <= m; ++j){
if(j) dp[i][j] |= dp[i+1][j-1];
if(j > 1) dp[i][j] |= dp[i+1][j-2];
}
}
} if(!dp[1][m]) puts("Impossible");
else{
int cnt = m;
for(int i = 1; i <= mid; ++i){
for(int j = 0; j < 26; ++j){
if(s[i] == 'a'+j || s[i+mid] == 'a'+j){
if(s[i] == s[i+mid]){
if(dp[i+1][cnt]) { ans[i] = ans[i+mid] = s[i]; break; }
}
else{
if(dp[i+1][cnt-1]){ ans[i] = ans[i+mid] = 'a' + j; --cnt; break; }
}
}
else{
if(dp[i+1][cnt-2]){ ans[i] = ans[i+mid] = 'a' + j; cnt -= 2; break; }
}
}
}
ans[n+1] = 0;
puts(ans+1);
}
}
return 0;
}