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//680K 32MS
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using
std::string;
using
std::vector;
#define DNA_Len 60
bool
BMSearch(const
char* s, const
char* t)
{
const
char *p = strstr(s, t);
if
(p) return
true;
else
return false;
}
int
main()
{
char
temp[DNA_Len+1];
vector<string> subStr[DNA_Len+1]; // 3-60
int
caseNum;
scanf("%d",&caseNum);
while
(caseNum-- > 0)
{
int
deqNum;//2 <= m <= 10
scanf("%d", &deqNum);
char
**p = new
char*[deqNum];
for
(int i=0; i<deqNum; ++i)
{
p[i] = new
char[DNA_Len+1];
memset(p[i], 0, DNA_Len+1);
scanf("%s", p[i]);
}
//1. 获取第一个串中 长度为3-60的所有子串 （小于3的输出no significant commonalities）
for
(int len=3; len<=60; ++len){
subStr[len].clear();
for
(int i=0; i<=60-len; ++i){
strncpy(&temp[0], p[0]+i, len);
temp[len] = 0;
subStr[len].push_back(temp);
}
}
//2. 子串由少到多 再deqNum个主串中查找， 如果都有该子串 则保存后查找下一个长度的子串，直到找不到
memset(temp, 0, DNA_Len+1);
bool
hasOneNotGot; //标记 deqNum个主串中，如果有一个没有找到那么就为true
for
(int subStrLen=3; subStrLen<=60; ++subStrLen){
for
(int subNum=0,count=subStr[subStrLen].size(); subNum<count; ++subNum){
hasOneNotGot = false;
for
(int strIdx=0 ; strIdx<deqNum; ++strIdx){
if
(! BMSearch(p[strIdx], subStr[subStrLen].at(subNum).c_str())){
hasOneNotGot = true;
break;
}
}
if
(! hasOneNotGot) { //找到了就退出，找下一个长度的串,没找到就继续
strcpy(temp, subStr[subStrLen].at(subNum).c_str());
break;
}
}
if
(hasOneNotGot){ //该长度的子串没有找到 那么temp是最长的子串 或temp为空
break;
}
}
if
(strlen(temp) >= 3){
//多个最长子串 按照字典排序
int
len = strlen(temp);
vector<string> multiString;
bool
searched;
for
(int i=0,count=subStr[len].size(); i<count; ++i)
{
searched = true;
for
(int strIdx=0 ; strIdx<deqNum; ++strIdx){
if
(! BMSearch(p[strIdx], subStr[len].at(i).c_str())){
searched = false;
break;
}
}
if
(searched) multiString.push_back(subStr[len].at(i));
}
strcpy(temp, multiString.at(0).c_str());
for
(int i=1,count=multiString.size(); i<count; ++i)
{
if
(strcmp(temp, multiString.at(i).c_str()) > 0){
strcpy(temp, multiString.at(i).c_str());
}
}
printf("%s\n", temp);
}
else
{
printf("no significant commonalities\n");
}
for
(int i=0; i<deqNum; ++i) delete
[](p[i]);
delete
[]p;
}
return
0;
}
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