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/* Dynamic Programming C implementation of LCS problem */
#include <bits/stdc++.h>
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 program to test above function */
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
int m = strlen(X);
int n = strlen(Y);
printf("Length of LCS is %d", lcs( X, Y, m, n ) );
return 0;
}
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#include <iostream>
#include <cstring>
using namespace std;
int f(string s1,string s2)
{
if(s1.length()==0||s2.length()==0)
return 0;
if(s1[0]==s2[0])
return 1+f(s1.substr(1),s2.substr(1));
else
return max(f(s1.substr(1),s2),f(s1,s2.substr(1)));
}
int main()
{
cout<<f("ABCD","ABD")<<endl;
return 0;
}
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int lcm(int a, int b)
{
// static variable
static int common = 0;
// add largest number
common += b;
// base case
if(common%a == 0) return common;
// general case
else return lcm(a, b);
}