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HDU 2577

2019-11-14 10:18:47
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How to Type

Time Limit: 2000/1000 MS (java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6440 Accepted Submission(s): 2909PRoblem DescriptionPirates 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.InputThe 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.OutputFor each test case, you must output the smallest times of typing the key to finish typing this string.Sample Input
3PiratesHDUacmHDUACMSample Output
888HintThe 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 8The string "HDUACM", can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8思路:dpa與dpb數(shù)組分別表示cap鍵開關(guān)狀態(tài),在完成第i個字母的最少步驟有兩種可能,在cap開和關(guān)狀態(tài),在第i-1個字母完成時,有兩種可能,一種是cap鍵開,一種是關(guān),每次取i-1狀態(tài)步驟的最小值,最后比較兩種狀態(tài)下,哪一種步驟最少。

#include<stdio.h>#include<string.h>char str[110];int dpa[110],dpb[110]; //dpa[110]表示燈亮,dpb[110]表示燈滅 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);//如果燈亮,按shift+字母,燈滅,按字母+cap 				dpb[i] = Min(dpa[i-1] + 2, dpb[i-1] + 1);//如果燈亮,按cap+字母,燈滅,按字母 			}			else if(str[i] >= 'A' && str[i] <= 'Z') {				 dpa[i] = Min(dpa[i-1] + 1, dpb[i-1] + 2);//如果燈亮,按字母,燈滅,按cap+字母 				 dpb[i] = Min(dpa[i-1] + 2, dpb[i-1] + 2);//如果燈亮,按字母+cap,燈滅,按shift字母 			}		}		printf("%d/n",Min(dpa[i-1] +1, dpb[i-1]));//燈亮著要關(guān)滅 	}	return 0;}


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