How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6440 Accepted Submission(s): 2909 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：dpa与dpb数组分别表示cap键开关状态，在完成第i个字母的最少步骤有两种可能，在cap开和关状态，在第i-1个字母完成时，有两种可能，一种是 cap键开，一种是关，每次取i-1状态步骤的最小值，最后比较两种状态下，哪一种步骤最少。

#include<stdio.h> #include<string.h> char str[110]; int dpa[110],dpb[110]; //dpa[110]表示灯亮，dpb[110]表示灯灭 int Min(int a, int b) { return a > b ? b : a; } int main() { int t,i; scanf("%d",&t); getchar(); dpa[0] = 1; dpb[0] = 0; while(t--) { scanf("%s",str + 1); for(i = 1; str[i]; i++) { if(str[i] >= 'a' && str[i] <= 'z') { dpa[i] = Min(dpa[i-1] + 2, dpb[i-1] + 2);//如果灯亮，按shift+字母，灯灭，按字母+cap dpb[i] = Min(dpa[i-1] + 2, dpb[i-1] + 1);//如果灯亮，按cap+字母，灯灭，按字母 } else if(str[i] >= 'A' && str[i] <= 'Z') { dpa[i] = Min(dpa[i-1] + 1, dpb[i-1] + 2);//如果灯亮，按字母，灯灭，按cap+字母 dpb[i] = Min(dpa[i-1] + 2, dpb[i-1] + 2);//如果灯亮，按字母+cap，灯灭，按shift字母 } } printf("%d\n",Min(dpa[i-1] +1, dpb[i-1]));//灯亮着要关灭 } return 0; }