广州大学第九届ACM B--再来一道

B - 再来一道

Time Limit: 2000/1000 MS (Java/Others)      Memory Limit: 128000/64000 KB (Java/Others) 

Submit Status

Problem Description

XiaoMing is so lazy that he was sleeping in his English lessons, so his English teacher punished him to calculate how many times 

that "sleep" and "sleepy" appear altogether in a given English string, and the string’s length is not greater than 1,000,000.

Input

The first line of the input contains an integer T(1<=T<=11) which means the number of test cases.

Each case contains a English string(which maybe contains lowercase, ‘,’, ’.’, and ’?’, but not ‘ (space)‘) in a single line.

Output

For each case, you need to print the answer in a single line.

Sample Input

5sleep.....ysl.eepxxsleepsleepyyyboringsleepppppppppppppppppppfifjfjo,hhhfijdfij.efejo?fdffdf

Sample Output

10310

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

/*

* this code is made by 1406100108

* Problem: 1353

* Verdict: Accepted

* Submission Date: 2015-04-14 14:29:13

* Time: 12 MS

* Memory: 2380 KB

*/

#include<iostream>

#include<cstring>

usingnamespace std;

 

intmain()

{

    chartext[1000001];

    unsigned inti,lg,T ,count;          //lg为字符串长

    cin>>T;

    while(T--)

    {

        count = 0;

        cin>>text;

        lg=strlen(text);

        for(i=0;i<lg;++i)

        {

            if(text[i]=='s')

                if(text[i+1]=='l')

                    if(text[i+2]=='e')

                        if(text[i+3]=='e')

                            if(text[i+4]=='p')

                            {

                                ++count;

                                if(text[i+5]=='y')

                                    ++count;

                            }

        }

        cout<<count<<endl;

    }

    return0;

}

这道题比较简单，主要就是寻找sleep和sleepy，注意在寻找到sleep时再后面可以再判断一下有没有y，有则累加。用这种方法破解有点暴力，缺乏简洁，这里还可用<string>

库里的字符串寻找函数，这样的话定义时应该定义成string   str。用直接寻找串内的子串。