HDU-2577-How to Type(模拟)

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7021    Accepted Submission(s): 3167

Problem Description

Pirates have finished developing the typing software. He called Cathy to test his typing software. She is good at thinking. After testing for several days, she finds that if she types a string by some ways, she will type the key at least. But she has a bad habit that if the caps lock is on, she must turn off it, after she finishes typing. Now she wants to know the smallest times of typing the key to finish typing a string.

 

Input

The first line is an integer t (t<=100), which is the number of test case in the input file. For each test case, there is only one string which consists of lowercase letter and upper case letter. The length of the string is at most 100.

 

Output

For each test case, you must output the smallest times of typing the key to finish typing this string.

 

Sample Input

3PiratesHDUacmHDUACM

 

Sample Output

888
Hint
The string “Pirates”, can type this way, Shift, p, i, r, a, t, e, s, the answer is 8.The string “HDUacm”, can type this way, Caps lock, h, d, u, Caps lock, a, c, m, the answer is 8The string "HDUACM", can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8
  

 flag表示capslock的状态  1表示打开  0为关闭  开始是和输入完毕都是关闭的关闭的  用plu记录shift和capslock的按键次数

当接下来输入的字母有连续n个跟capslock状态不同时  分析可只  只有n=1时适合用shift键  

如flag=1 n=1  输入a时  shift+a=2   而capslock+a+capslock=3

n>=2  如输入ab是  shift+a+shift+b=4   capslock+a+b+capslock=4

所以每次输入的字母如capslock状态不同时  就看有几个连续状态不同的字母  然后再选择用什么键

注意有一个例外当最后只有一个字母小写 capslock打开时 也应该用capslock  因为最后要换位小写状态

#include<cstdio>#include<cstring>#include<string>using namespace std;int main(){    int t;    char ss[105];    int pul;    bool falg;    scanf("%d",&t);    while(t--){        falg=false;        pul=0;        scanf("%s",ss+1);        int l=strlen(ss+1);        //printf("%d\n",l);        int t;        for(int i=1;i<=l;i++){            t=0;            if(falg) while(islower(ss[i])){t++;i++;}            else while(isupper(ss[i])){t++;i++;}            if(t>1||(i==l+1&&islower(ss[l])&&t==1)){                falg=!falg;                i--;pul++;            }            else if(t==1){                i--;                pul++;            }        }        if(falg) pul++;        printf("%d\n",l+pul);    }    return 0;}