HDOJ --- 2577

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3261    Accepted Submission(s): 1509


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

思路：openclock[i]记录到i位置打开大写键时的最小按键次数，closeclock[i]记录到i位置时不打开大写键时最小按键次数，详见代码。

#include<ctype.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#define MAX 111
using namespace std;
int closeclock[MAX], openclock[MAX];
char str[MAX];
int main(){
    int c, len;
    /* freopen("in.c", "r", stdin); */
    scanf("%d", &c);
    while(c--){
        memset(str, 0, sizeof(str));
        scanf("%s", str);
        len = strlen(str);
        openclock[0] = 1;
        for(int i = len;i > 0;i --) str[i] = str[i-1];
        for(int i = 1;i <= len;i ++){
            if(islower(str[i])){
                openclock[i] = min(openclock[i-1]+2, closeclock[i-1]+2);
                closeclock[i] = min(openclock[i-1]+2, closeclock[i-1]+1);
            }else{
                openclock[i] = min(openclock[i-1]+1, closeclock[i-1]+2);
                closeclock[i] = min(openclock[i-1]+2, closeclock[i-1]+2);
            }
        }
        openclock[len]++;
        printf("%d\n", min(openclock[len], closeclock[len]));
    }
    return 0;
}