# hdu 5903 Square Distance(dp)

Problem Description

A string is called a square string if it can be obtained by concatenating two copies of the same string. For example, "abab", "aa" are square strings, while "aaa", "abba" are not.

Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.

Peter has a string s=s1s2...sn of even length. He wants to find a lexicographically smallest square string t=t1t2...tn that the hamming distance between s and t is exact m. In addition, both s and t should consist only of lowercase English letters.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains two integers n and m (1≤n≤1000,0≤m≤n,n is even) -- the length of the string and the hamming distance. The second line contains the string s.

Output

For each test case, if there is no such square string, output "Impossible" (without the quotes). Otherwise, output the lexicographically smallest square string.

dp[i][count]表示取到i位有count个位置不一样是否成立。dp[i][count]+=dp[i+1][count-gg](gg表示要变的位置数）

（1）s[i]==s[i+n/2]，这个时候要么不变位置要么就变2次。

（2）s[i]!=s[i+n/2]，这是要么变2次要么变1次。

#include <iostream>
#include <cstring>
using namespace std;
char s[1010];
int dp[1010][1010];
int main()
{
int t;
cin >> t;
while(t--) {
int n , m;
cin >> n >> m;
cin >> (s + 1);
int count = m;
memset(dp , 0 , sizeof(dp));
dp[n / 2 + 1][0] = 1;
for(int i = n / 2 ; i >= 1 ; i--) {
if(s[i] == s[i + n / 2]) {
for(int j = 0 ; j <= m ; j++) {
dp[i][j] += dp[i + 1][j];
}
for(int j = 2 ; j <= m ; j++) {
dp[i][j] += dp[i + 1][j - 2];
}
}
else {
for(int j = 1 ; j <= m ; j++) {
dp[i][j] += dp[i + 1][j - 1];
}
for(int j = 2 ; j <= m ; j++) {
dp[i][j] += dp[i + 1][j - 2];
}
}
}
if(!dp[1][m]) {
cout << "Impossible" << endl;
continue;
}
for(int i = 1 ; i <= n / 2 ; i++) {
for(int j = 0 ; j < 26 ; j++) {
int temp = (s[i] != j + ‘a‘) + (s[i + n / 2] != j + ‘a‘);
if(dp[i + 1][m - temp]) {
s[i] = s[i + n / 2] = j + ‘a‘;
m -= temp;
break;
}
}
}
cout << s + 1 << endl;
}
return 0;
}


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