Description

Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

Certainly, we would like to minimize the number of all possible operations.

Illustration

A G T A A G T * A G G C

| | |       |   |   | |

A G T * C * T G A C G C

Deletion: * in the bottom line
Insertion: * in the top line
Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C

|  |  |        |     |     |  |

A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥ m.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y.

Input

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string x into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4

题意 给你两个DNA序列 求第一个第一个序列至少经过多次删除 、替换 或添加碱基得到第二个序列 其实分析一下可以发现 只要求出两个序列的最长公共子序列 这部分就可以不动了 然后较长序列的长度减去最长公共子序列的长度就是答案了

#include
  
   
#include
   
     #include
    
      using namespace std; const int N = 1005; int la, lb, d[N][N]; char a[N], b[N]; void lcs() { memset (d, 0, sizeof (d)); for (int i = 1; i <= la; ++i) for (int j = 1; j <= lb; ++j) if (a[i] == b[j]) d[i][j] = d[i - 1][j - 1] + 1; else d[i][j] = max (d[i - 1][j], d[i][j - 1]); } int main() { while (~scanf ("%d%s%d%s", &la, a + 1, &lb, b + 1)) { lcs(); printf ("%d\n", max (la, lb) - d[la][lb]); } return 0; }