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原创最新推荐文章于 2026-07-18 11:37:36 发布 8.6k 阅读 AI 写同款· GEO 优化 ›

[资源分享]一个类似LOL的UNITY分享 3D项目来源参考资料1，原文 5V5地图模型上传了,下面链接: 链接：http://pan.baidu.com/s/1mibndKc 密码：xgwy 参考资料： 1. 【模仿】英雄联盟类游戏不断更新 2. 3. 阅读详情

LOL

Time Limit: 1000ms, Special Time Limit:2500ms,Memory Limit:65366KB

Problem description

Your friend Andreas has a very good sense of humour. In fact, it is so good that he even enjoys to change words so that they will contain the substring lol. For example,during the 2010 FIFA World Cup in soccer he had much fun changing the word fotball to fotbalol. In order to be more ecient in his word plays, he has asked you to make a computer program that nds the minimum number of changes needed in order to transform a string into a new string containing lol as a substring. There are three legal ways to change a string: delete a character, insert a new character and replace an existing character with a new one.

Input

The first line of the input consists of a single number T, the number of test cases. Then follows T lines, containing a string S consisting of lower case letters between a and z only.

Output

For each string, output on its own line the minimum number of changes needed in order to obtain a string containing lol as a substring.

Sample Input

4
fotball
sopp
ingenting
spillolje

Sample Output

Judge Tips

0 < T <= 100 0 < |S| <= 50 (That is, the maximal string length is 50.)

Problem Source

IDIOPEN 2011

题目思路：只要查找字符子串，再处理就OK的了

program：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
char ch[55];
int main()
{
int test;
cin>>test;
while(test--)
{
   cin>>ch;
   if(strlen(ch)==1)
      {
        if(ch[0]=='l'||ch[0]=='o')
          {
           cout<<2<<endl;
            continue;
          }
        else
          {
             cout<<3<<endl;
             continue;
          }

      }
   char *p;
   char a[5]={"lol"},b[5]={"lo"},c[5]={"ol"},d='l',e='o',f[5]={"ll"};
   p=strstr(ch,a);
   if(p)
     {
         cout<<0<<endl;
         //cout<<"---------1"<<endl;
         continue;
     }
   p=strstr(ch,b);//////
   if(p)
     {
         cout<<1<<endl;
         //cout<<"---------2"<<endl;
         continue;
     }
   p=strstr(ch,c);//////
   if(p)
     {
         cout<<1<<endl;
         //cout<<"---------3"<<endl;
         continue;
     }
   p=strstr(ch,f);//////
   if(p)
     {
         cout<<1<<endl;
         //cout<<"---------4"<<endl;
         continue;
     }
   for(int i=0;i<(strlen(ch)-2);i++)
    {
       if(ch[i]=='l'&&ch[i+2]=='l')
        {
           cout<<1<<endl;
          // cout<<"---------5"<<endl;
           goto end;
        }
    }

   p=strchr(ch,d);/////
   if(p)
     {
         cout<<2<<endl;
         //cout<<"---------6"<<endl;
         continue;
     }
   p=strchr(ch,e);/////
   if(p)
     {
         cout<<2<<endl;
         //cout<<"---------7"<<endl;
         continue;
     }
   cout<<3<<endl;

end:;
}
//system("pause");
return 0;
}