Square Distance (构造)

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本文介绍了一个算法问题，目标是在给定字符串和特定哈明距离条件下找到字典序最小的平方字符串。文章详细解释了问题背景、输入输出要求，并提供了一段示例代码实现。

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Square Distance

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 312 Accepted Submission(s): 102

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

tis 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

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

Output

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

Sample Input

3
4 1
abcd
4 2
abcd
4 2
abab

Sample Output

Impossible
abab
aaaa

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 10010
#define MOD 1000000007
const int mod = 2520;

int gcd(int a, int b)
{
while(b)
{
int t = a % b;
a = b;
b = t;
}
return a;
}

int T,n,m;

int sum[550];

char st[1200];
char re[1200];

int main()
{
scanf("%d",&T);

while(T--)
{
scanf("%d%d",&n,&m);

scanf("%s",&st);

int tt=0;
memset(sum,0,sizeof(sum));

for(int i=0;i<n/2;i++)
{
if(st[i]!=st[i+n/2]) tt++;

sum[i] = tt;
}

if(tt>m) {printf("Impossible\n");continue;}

int ff=0;
bool flag=true;
for(int i=0;i<n/2;i++)
{
int tmp=0,uu,oo;

re[i] = 'a';

if(re[i]!=st[i]) tmp++;
if(re[i]!=st[i+n/2]) tmp++;

uu=ff+tmp+sum[n/2-1] - sum[i];
oo=ff+tmp+ (n/2-1 - i)*2;
if(uu<=m&&oo>=m)
{
ff+=tmp;
continue;
}

tmp=0;
re[i] = 'b';

if(re[i]!=st[i]) tmp++;
if(re[i]!=st[i+n/2]) tmp++;

uu=ff+tmp+sum[n/2-1] - sum[i];
oo=ff+tmp+ (n/2-1 - i)*2;
if(uu<=m&&oo>=m)
{
ff+=tmp;
continue;
}

tmp=0;
re[i] = 'c';

if(re[i]!=st[i]) tmp++;
if(re[i]!=st[i+n/2]) tmp++;

uu=ff+tmp+sum[n/2-1] - sum[i];
oo=ff+tmp+ (n/2-1 - i)*2;
if(uu<=m&&oo>=m)
{
ff+=tmp;
continue;
}

tmp=0;
re[i] =min(st[i],st[i+n/2]);

if(re[i]!=st[i]) tmp++;
if(re[i]!=st[i+n/2]) tmp++;

uu=ff+tmp+sum[n/2-1] - sum[i];
oo=ff+tmp+ (n/2-1 - i)*2;
if(uu<=m&&oo>=m)
{
ff+=tmp;
continue;
}

flag=false;
// printf("hkfghfkhg\n");
}

if(flag==false)
{
printf("Impossible\n");continue;
}

for(int i=0;i<n/2;i++)
{
printf("%c",re[i]);
}
for(int i=0;i<n/2;i++)
{
printf("%c",re[i]);
}
printf("\n");

}

return 0;
}