CodeForces 156A Message(暴力)

A. Message

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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;
}