CQUOJ 2475 - How to Type

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

 

/*
2016年5月16日23:32:06
题目大意： 给你一些字符串a~z||A~Z，求能把它们都打出来的最少次数. 
如果我们需要打a的话
    如果开始Cap键没开，直接打, 只需1步, Cap键仍是关的. 
    如果开始Cap键开了, 可以先按Cap键,再按a,需要2步,不过此时Cap键关了. 
                  或者也可以按下shift，再按下a，需要两步, 不过Cap键仍是开的. 
打A的话同样情况分析 

还有就是 
    一开始Cap键是关的, 且最后完成后需要使Cap键关闭。 
    
    dp[i][0] 表示打第i个字符,且此时Cap键是关的所需最小次数
    dp[i][1] 表示打第i个字符,且此时Cap键是开的所需最小次数
*/


# include <iostream>
# include <algorithm>
# include <cstdio>
# include <cstring>
# include <queue>
# include <vector>
# include <cmath>
# include <map>
# include <set>
# define pi acos(-1)
# define LL long long
# define INF 0x3f3f3f3f
using namespace std;

char s[105];
int dp[105][2];
int main(void)
{
    int len, i, ans, tes;
    scanf("%d", &tes);
    while (tes--){
        scanf("%s", &s);
        len = strlen(s);
        dp[0][0] = 0;
        dp[0][1] = 1;
        for (i = 0; i < len; i++){
            if (s[i]<='z' && s[i]>='a'){
                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);
            }    
            else {
                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);
            }
        }
        ans = min(dp[len][0], dp[len][1]+1);
        cout<<ans<<endl;
    }
    
    return 0;
}