DP:How To Type

How to Type Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2577

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
 

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char str[110];
int dp[110][2]; //打i号字母前按下的最小次数，j=0表示caplocks按了，1表示没按

int main(){
    int t, l, i;
    cin >> t;
    while(t --){
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 0;
        dp[0][1] = 1;
        scanf("%s", str);
        l = strlen(str);
        for(i = 0; i <= l-1; i ++){ //一共0到i-1号字母
            if(str[i] >= 'A' && str[i] <= 'Z'){
                dp[i+1][0] = min(dp[i][0] + 2, dp[i][1] + 2);
                dp[i+1][1] = min(dp[i][0] + 2, dp[i][1] + 1);
            }
            else{
                dp[i+1][0] = min(dp[i][0] + 1, dp[i][1] + 2);
                dp[i+1][1] = min(dp[i][0] + 2, dp[i][1] + 2);
            }
        }
        cout << min(dp[l][0], dp[l][1] + 1) << endl;
    }
    return 0;
}