hdu_2577 How to Type

How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6213 Accepted Submission(s): 2806

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

3PiratesHDUacmHDUACM

Sample Output

888
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 8The string "HDUACM", can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8
简单dp
dp[i][0]表示输入第i个字母后Caps键未开启
dp[i][1]表示输入第i个字母后Caps键开启
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 1000010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;char a[110];int dp[110][3];int n;int main(){    scanf("%d", &n);    while(n--){        memset(dp, 0, sizeof(dp));        scanf("%s", a);        int len = strlen(a);        if(a[0] >= 'a' && a[0] <= 'z'){            dp[0][0] = 1;            dp[0][1] = 2;        }        else if(a[0] >= 'A' && a[0] <= 'Z'){            dp[0][0] = 2;            dp[0][1] = 2;        }        for(int i = 1; a[i] != '\0'; i++){            if(a[i] >= 'a' && a[i] <= 'z'){                dp[i][0] = min(dp[i-1][0] + 1, dp[i-1][1] + 2);                dp[i][1] = min(dp[i-1][0] + 2, dp[i-1][1] + 2);            }            else if(a[i] >= 'A' && a[i] <= 'Z'){                dp[i][0] = min(dp[i-1][0] + 2, dp[i-1][1] + 2);                dp[i][1] = min(dp[i-1][0] + 2, dp[i-1][1] + 1);            }        }        printf("%d\n", min(dp[len-1][1] + 1, dp[len-1][0]));    }    return 0;}