题目：
In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (maybe except for Luxembourg). To enforce that Germany will fulfill the criteria, our government has so many wonderful options (raise taxes, sell stocks, revalue the gold reserves,...) that it is really hard to choose what to do。
Therefore the German government requires a program for the following task:
Two politicians each enter their proposal of what to do. The computer then outputs the longest common subsequence of words that occurs in both proposals. As you can see, this is a totally fair compromise (after all, a common sequence of words is something what both people have in mind).
Your country needs this program, so your job is to write it for us.
Input Specification
The input file will contain several test cases.
Each test case consists of two texts. Each text is given as a sequence of lower-case words, separated by whitespace, but with no punctuation. Words will be less than 30 characters long. Both texts will contain less than 100 words and will be terminated by a line containing a single ‘#‘.
Input is terminated by end of file.
Output Specification
For each test case, print the longest common subsequence of words occuring in the two texts. If there is more than one such sequence, any one is acceptable. Separate the words by one blank. After the last word, output a newline character.
Sample Input
die einkommen der landwirte sind fuer die abgeordneten ein buch mit sieben siegeln um dem abzuhelfen muessen dringend alle subventionsgesetze verbessert werden # die steuern auf vermoegen und einkommen sollten nach meinung der abgeordneten nachdruecklich erhoben werden dazu muessen die kontrollbefugnisse der finanzbehoerden dringend verbessert werden #
Sample Output
die einkommen der abgeordneten muessen dringend verbessert werden解析：简单的LCS不过加了路径的打印，和01背包的路径打印有所不同 这道题的输入方法值得学习，输入一大串字符串直到出现字符“#”，为止。 可以用一个二维数组，先输入第一个单词，判断首字母是不是“#”。
1 #include<cstdio>
2 #include<cstring>
3 const int maxn=110;
4 char str1[maxn][32],str2[maxn][32];
5 int dp[maxn][maxn],path[maxn][maxn],flag;
6 void print(int x,int y)
7 {
8 if(!x||!y)
9 return;
10 if(path[x][y]==1)
11 {
12 print(x-1,y-1);
13 if(flag)
14 printf(" ");
15 else
16 flag=1;
17 printf("%s",str1[x]);
18 }
19 else if(path[x][y]==0)
20 print(x-1,y);
21 else
22 print(x,y-1);
23 }
24 int main()
25 {
26 int num1,num2;
27 while(scanf("%s",str1[1])==1)
28 {
29 num1=1;
30 num2=0;
31 if(str1[1][0]!=‘#‘)
32 {
33 for(int i=2;;i++)
34 {
35 scanf("%s",str1[i]);
36 if(str1[i][0]==‘#‘) break;
37 num1++;
38 }
39 }
40 for(int i=1;;i++)
41 {
42 scanf("%s",str2[i]);
43 if(str2[i][0]==‘#‘) break;
44 num2++;
45 }
46 memset(dp,0,sizeof(dp));
47 for(int i=1;i<=num1;i++)
48 {
49 for(int j=1;j<=num2;j++)
50 {
51 if(!strcmp(str1[i],str2[j]))
52 {
53 dp[i][j]=dp[i-1][j-1]+1;
54 path[i][j]=1;
55 }
56 else if(dp[i-1][j]>dp[i][j-1])
57 {
58 dp[i][j]=dp[i-1][j];
59 path[i][j]=0;
60 }
61 else
62 {
63 dp[i][j]=dp[i][j-1];
64 path[i][j]=-1;
65 }
66 }
67 }
68 flag=0;
69 print(num1,num2);
70 printf("\n");
71 }
72 return 0;
73 }