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  • 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
    ",ans);
    
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/dacc123/p/8228643.html
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