Description：

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

Input：

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output：

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.  

Sample Input：

abcfbc abfcab

programming contest

abcd mnp

Sample Output：

4

2

0

解题思路：

题目的意思就是给你两个字符串，让你求最长公共子序列，就是一个标准的DP，直接套用公式就可以了！！！

程序代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 1100
char a[N],b[N];
int dp[N][N];
int main()
{
  while(cin>>a>>b)
  {
    int len1=strlen(a);
    int len2=strlen(b);
    for(int i=1;i<=len1;i++)
    {
      for(int j=1;j<=len2;j++)
      {
        if(a[i-1]==b[j-1])
          dp[i][j]=dp[i-1][j-1]+1;
        else
          dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
      }
    }
    cout<<dp[len1][len2]<<endl;
  }
  return 0;
}