/* Dynamic Programming C++ implementation of LCS problem */
#include<bits/stdc++.h>
#include <fstream>
using namespace std;
ifstream fin("cmlsc.in");
ofstream fout("cmlsc.out");

int max(int a, int b);

/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
    int L[m + 1][n + 1];
    int i, j;

    /* Following steps build L[m+1][n+1] in
       bottom up fashion. Note that L[i][j]
       contains length of LCS of X[0..i-1]
       and Y[0..j-1] */
    for (i = 0; i <= m; i++)
    {
        for (j = 0; j <= n; j++)
        {
        if (i == 0 || j == 0)
            L[i][j] = 0;

        else if (X[i - 1] == Y[j - 1])
            L[i][j] = L[i - 1][j - 1] + 1;

        else
            L[i][j] = max(L[i - 1][j], L[i][j - 1]);
        }
    }

    /* L[m][n] contains length of LCS
    for X[0..n-1] and Y[0..m-1] */
    return L[m][n];
}

/* Utility function to get max of 2 integers */
int max(int a, int b)
{
    return (a > b)? a : b;
}

// Driver Code
int main()
{
    char X[] = "AGGTAB";
    char Y[] = "GXTXAYB";

    int m = strlen(X);
    int n = strlen(Y);

    fout << "Length of LCS is "
         << lcs( X, Y, m, n );

    return 0;
}