Time Limit: 1000MS

Memory Limit: 10000K

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

問題分析

題意：求倆個字串的最長公共子串的長度
一道經典dp。
在解決這一類的問題時，我們所需要做的肯定是
⒈找子問題
⒉確定狀態
⒊找出狀態轉移方程
那怎麼找子問題呢？題意要求兩個字串的最長公共子串，不如試著去求s1左邊0->s1.length-1和s2左邊0->s2.length-1的的最長公共部分。也就是在0->s1.length-1和0->s2.length-1中求maxLength(i,j)，因此，maxLength(i,j)就是本題的狀態。所以我們可以通過去比較s1[i-1]和s2[j-1],來遞推maxLength(i,j)。所以本題的狀態方程為
if(s1[i-1]==s2[j-1])

好，下面上ACcode（^_^）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
const int inf = 0x3f3f3f;
const int N = 500;
using namespace std;

int dp[N][N];

int main()
{
    char a[N],b[N];
    while(~scanf(" %s %s",a ,b))
    {
        int a1 = strlen(a);
        int b1 = strlen(b);
        for(int i = 1; i <= a1; i++)
        {
            for(int j = 1; j <= b1; ++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[a1][b1]<<endl;
    }
    return 0;
}