CodeForces 156A Message(暴力)-优快云博客

本文介绍了一个算法问题：从给定字符串中选择一个子串，通过最少的操作（插入、删除或替换字符）转换为另一个固定字符串。文章提供了一个C++实现示例。

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Dr. Moriarty is about to send a message to Sherlock Holmes. He has a string s.

String p is called a substring of string s if you can read it starting from some position in the string s. For example, string "aba" has six substrings: "a", "b", "a", "ab", "ba", "aba".

Dr. Moriarty plans to take string s and cut out some substring from it, let's call it t. Then he needs to change the substring t zero or more times. As a result, he should obtain a fixed string u (which is the string that should be sent to Sherlock Holmes). One change is defined as making one of the following actions:

Insert one letter to any end of the string.
Delete one letter from any end of the string.
Change one letter into any other one.

Moriarty is very smart and after he chooses some substring t, he always makes the minimal number of changes to obtain u.

Help Moriarty choose the best substring t from all substrings of the string s. The substring t should minimize the number of changes Moriarty should make to obtain the string u from it.

Input

The first line contains a non-empty string s, consisting of lowercase Latin letters. The second line contains a non-empty string u, consisting of lowercase Latin letters. The lengths of both strings are in the range from 1 to 2000, inclusive.

Output

Print the only integer — the minimum number of changes that Dr. Moriarty has to make with the string that you choose.

Examples

input
aaaaa
aaa

output
0

input
abcabc
bcd

output
1

input
abcdef
klmnopq

output

7



暴力

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
string a;
string b;
int main()
{
    cin>>a>>b;



    int len1=a.length();
    int len2=b.length();
    int ans=10000000;
    int len=len2;
    if(len1<len2)
       {
          swap(a,b);
          swap(len1,len2);
       }

    for(int i=1;i<len2;i++)
    {
           int num=0;
        for(int k=i,j=0;k<len2&&j<len1;k++,j++)
        {
           if(b[k]==a[j])
            num++;
        }
        ans=min(ans,len-num);
    }

    for(int i=0;i<len1;i++)
    {
        int num=0;
        for(int k=i,j=0;j<len2&&k<len1;j++,k++)
        {
            if(a[k]==b[j])
                num++;
        }

        ans=min(ans,len-num);
    }
    printf("%d\n",ans);

    return 0;
}