HDU OJ How to Type

How to Type

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 45   Accepted Submission(s) : 29

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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

<div style='font-family:Times New Roman;font-size:14px;background-color:F4FBFF;border:#B7CBFF 1px dashed;padding:6px'><div style='font-family:Arial;font-weight:bold;color:#7CA9ED;border-bottom:#B7CBFF 1px dashed'><i>Hint</i></div>
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
</div>

分析:输入以字符串，求最少需要敲多少下键盘才能完成输入！dp[i][0],dp[i][1]分别记录灯亮时输入第i个字符所敲键盘的次数，灯灭时输入第i个字符时所敲键盘的次数！状态转移方程参见代码！

代码如下：

#include<iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int main()
{
    int i,n;
    char str[105];
    int len;
    int dp[105][2];
    int Min;
    cin>>n;
    while(n--)
    {
        cin>>str+1;
        memset(dp,0,sizeof(dp));
        dp[0][1]=1;
        len=strlen(str+1);
        for(i=1;i<=len;i++)
        {
            if(str[i]>='a' && str[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
            {
                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);
            }
        }
        dp[len][1]++;
        Min=dp[len][0]<dp[len][1]?dp[len][0]:dp[len][1];
        cout<<Min<<endl;
    }
    return 0;
}