求出最長子序列，且不包含一個子字串
DP 難題，要在 LCS 上面多開一維的空間
用來記錄病毒字串的生長，所以需要用 KMP
首次深入學習了
//
// GGGGGGGGGGGGG CCCCCCCCCCCCC AAA
// GGG::::::::::::G CCC::::::::::::C A:::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::::A
// G:::::G GGGGGG C:::::C CCCCCC A:::::::::A
//G:::::G C:::::C A:::::A:::::A
//G:::::G C:::::C A:::::A A:::::A
//G:::::G GGGGGGGGGGC:::::C A:::::A A:::::A
//G:::::G G::::::::GC:::::C A:::::A A:::::A
//G:::::G GGGGG::::GC:::::C A:::::AAAAAAAAA:::::A
//G:::::G G::::GC:::::C A:::::::::::::::::::::A
// G:::::G G::::G C:::::C CCCCCC A:::::AAAAAAAAAAAAA:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::A A:::::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A A:::::A
// GGG::::::GGG:::G CCC::::::::::::C A:::::A A:::::A
// GGGGGG GGGG CCCCCCCCCCCCCAAAAAAA AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 30005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-6
#define X first
#define Y second
clock\_t start\_time=clock(), end\_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
char virus\[105\],s1\[105\],s2\[105\];
int dp\[128\]\[128\]\[128\];
bool vis\[128\]\[128\]\[128\];
int acpos\[105\]\[128\];
bool checkac(int pos,int i,char x){
if(virus\[pos\]!=x){
return false;
}
for(int j=pos-1;j>=0;j--){
i--;
if(virus\[j\]!=virus\[i\])return false;
}
return true;
}
void initac(){
for(int i=0;virus\[i\];i++){
for(int j='A';j<='Z';j++){
for(int k=i;k>=0;k--){
if(checkac(k,i,j)){
acpos\[i\]\[j\]=k+1;
break;
}
}
}
}
}
int dfs(int pos1,int pos2,int pos3){
if(!virus\[pos3\])return -oo;
if(vis\[pos1\]\[pos2\]\[pos3\])return dp\[pos1\]\[pos2\]\[pos3\];
vis\[pos1\]\[pos2\]\[pos3\]=1;
if(s1\[pos1\]==0||s2\[pos2\]==0){
return 0;
}
if(s1\[pos1\]==s2\[pos2\]){
dp\[pos1\]\[pos2\]\[pos3\]=dfs(pos1+1,pos2+1,acpos\[pos3\]\[s1\[pos1\]\])+1;
}
dp\[pos1\]\[pos2\]\[pos3\]=max3(dp\[pos1\]\[pos2\]\[pos3\],dfs(pos1+1,pos2,pos3),dfs(pos1,pos2+1,pos3));
return dp\[pos1\]\[pos2\]\[pos3\];
}
void recon(int pos1,int pos2,int pos3){
if(!virus\[pos3\])return;
if(s1\[pos1\]==0||s2\[pos2\]==0)return ;
// debug("%d %d %d\\n",pos1,pos2,pos3);
if(dp\[pos1\]\[pos2\]\[pos3\]==dp\[pos1+1\]\[pos2\]\[pos3\]){
recon(pos1+1,pos2,pos3);
}else if(dp\[pos1\]\[pos2\]\[pos3\]==dp\[pos1\]\[pos2+1\]\[pos3\]){
recon(pos1,pos2+1,pos3);
}else{
printf("%c",s1\[pos1\]);
recon(pos1+1,pos2+1,acpos\[pos3\]\[s1\[pos1\]\]);
}
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%s %s %s",s1,s2,virus)){
initac();
Set(vis,0);
if(dfs(0,0,0)==0){
puts("0");
}else{
// printf("%d\\n",dfs(0,0,0));
recon(0,0,0);
puts("");
}
}
}