hdu 2577 How to Type (dp)

How to Type Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5245    Accepted Submission(s): 2347 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 题意：要求一个字符串输入，按键盘的最少次数。有Caps Lock和Shift两种转换大小写输入的方式 思路：用dpa与dpb数组分别记录Caps Lock的开关状态 #include <iostream> #include <cstring> #include <algorithm> #include <cstdlib> using namespace std; int dpa[105];//caps开的状态 int dpb[105];//关的状态 char str[105]; int main() { int t, i, j; scanf("%d", &t); while(t --) { scanf("%s", str + 1); int len = strlen(str + 1); dpa[0] = 0; dpb[0] = 1; // printf("%d\n",len); // printf("%c\n", str[1]); for(i = 1; i <= len ; i ++) { if(str[i] >= 'a' && str[i] <= 'z') { dpa[i] = min(dpa[i - 1] + 1, dpb[i - 1] + 2); dpb[i] = min(dpa[i - 1] + 2, dpb[i - 1] + 2); } else { dpa[i] = min(dpa[i - 1] + 2, dpb[i - 1] + 2); dpb[i] = min(dpa[i - 1] + 2, dpb[i - 1] + 1); } } printf("%d\n", min(dpa[len], dpb[len] + 1)); } }