How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

3
Pirates
HDUacm
HDUACM

Sample Output

8
8
8

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 8
The string "HDUACM", can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8

Author

Dellenge

Source

HDU 2009-5 Programming Contest

#include<stdio.h>
#include<string.h>
char s[110];
int da_ju(int i){
    if((s[i]>='A')&&(s[i]<='Z')) return 1;
    return 0;
}
int caps=0;
int fun(int i){
    if(caps){
          if(da_ju(i)) return 1;
        else {
           if(s[i+1]=='\0') {  //s[i] 是最后一个大写字母；
            caps=0;
            return 1;
           }
           else if(da_ju(i+1)) return 2;  //s[i] 是小写字母，而后跟的是大写字母，此时执行： shift+s[i]本身；
           else if(!da_ju(i+1)){     //s[i]是小写字母，且紧跟在后的也是小写字母，此时要关闭 caps；
                caps=0;
                return 1;
            }
        }
    }
    else{//caps未开
        if(da_ju(i)&&da_ju(i+1)) {  //caps未开，s[i]是大写字母，且紧跟在后的s[i+1] 也是大写字母，此时要打开caps；
            caps=1;
            return 3;    // caps的关闭操作始终归于使得caps打开的那个大写字母；
			  //实际上caps并未关闭，但它的关闭操作的按键数已经在打开时的同时就被计
			//数在内了，以后再关闭caps时，只需要修改caps的值即可（即caps=0），而不需要再计算操作数 ！！！！
        }
        else if(da_ju(i)){  // s[i]是大写，而s[i+1]是小写，执行： shift+s[i]
            return 2;
        }
        else return 1;  //s[i]是小写字母，s[i+1]也是小写字母
    }  

}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        caps=0;
        scanf("%s",s);
        int len=strlen(s);
            s[len]='\0';
        int res=0;
        for(int i=0;i<len;++i){
            res+=fun(i);
        }
        printf("%d\n",res);
    }
    return 0;
}  

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