How to Type
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7130 Accepted Submission(s): 3222
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
题意：
给定一串字母，问最少需要按多少次键盘将这些字母打印到屏幕上，键盘最开始的时候是小写锁定的，最后也要回归到小写锁定。
分析：
在小（大）写锁定下输入小（大）写字母肯定不需要进行考虑了，关键就是在小（大）写锁定下输入大（小）写字母，如果连续的只有一个的话，就用Shift键只需要多按一次，如果两个或者两个以上的话，就转换大小写锁定。
用 dp，贪心思想去写 相比模拟去写会简单很多。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int dpp[120],dpd[120];
char str[120];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s",str+1);
int i;
dpp[0]=0;//小写
dpd[0]=1;//大写
for(i=1;str[i];i++)
{
if(str[i]>='a'&&str[i]<='z')
{
dpp[i]=min(dpp[i-1]+1,dpd[i-1]+2);//若灯亮，则直接按字母；若灯灭，则按字母再开灯
dpd[i]=min(dpp[i-1]+2,dpd[i-1]+2);//若灯亮，则要先按字母再关灯；若灯灭，则按shift+字母
}
else if(str[i]>='A'&&str[i]<='Z')
{
dpp[i]=min(dpp[i-1]+2,dpd[i-1]+2);//若灯亮，则要先关灯再按字母；若灯灭，则字母
dpd[i]=min(dpp[i-1]+2,dpd[i-1]+1);//若灯亮，按字母；若灯灭，则按shift+字母
}
}
printf("%d
",min(dpp[i-1],dpd[i-1]+1));//最后要关灯，dpb要+1
}
return 0;
}