HDU 2577 How to Type (字符串处理)

题目链接

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:
给定一串字母,问最少需要按多少次键盘将这些字母打印到屏幕上,键盘最开始的时候是小写锁定的,最后也要回归到小写锁定。

分析:
在小(大)写锁定下输入小(大)写字母肯定不需要进行考虑了,关键就是在小(大)写锁定下输入大(小)写字母,如果连续的只有一个的话,就用Shift键只需要多按一次,如果两个或者两个以上的话,就转换大小写锁定。

代码:

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char ch[109];
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf(" %s",ch);
        int k=strlen(ch);
        int i=0;
        int ans=0;
        int flag=0;///Caps lock键有没有被按下,0表示小写锁定,1表示大写锁定
        while(i<k)
        {
            if(flag==0&&ch[i]>='a'&&ch[i]<='z')///小写锁定的情况下是小写字母
            {
                i++;
                ans++;
            }
            if(ch[i]>='A'&&ch[i]<='Z'&&flag==1)///大写锁定的情况下是大写字母
            {
                i++;
                ans++;
            }
            if(flag==0&&ch[i]>='A'&&ch[i]<='Z')///小写锁定的情况下是大写字母
            {
                if(i+1<k)
                {
                    if(ch[i+1]>='A'&&ch[i+1]<='Z')///下一个还是大写,就可以换成大写锁定
                    {
                        flag=1;
                        ans+=2;

                    }
                    else///否则直接用Shift键就行
                    {
                        ans+=2;

                    }
                }
                else///已经是最后一个字母了,肯定就是用shift键
                    ans+=2;
                i++;

            }
            if(flag==1&&ch[i]>='a'&&ch[i]<='z')///大写锁定的情况下是小写字母
            {
                if(i+1<k)
                {
                    if(ch[i+1]>='a'&&ch[i+1]<='z')///下一个还是小写,就可以换成小写锁定
                    {
                        flag=0;
                        ans+=2;
                    }
                    else///下一个字母大写的话,用shift键
                    {
                        ans+=2;
                    }
                }
                else///已经是左后一个字母了
                {
                    ans+=2;
                    flag=0;///肯定就回归小写锁定了
                }
                i++;
            }
        }
        if(flag==1)///最后害的看一下是不是大写锁定
            ans+=1;
        printf("%d
",ans);
    }
    return 0;
}

原文地址:https://www.cnblogs.com/cmmdc/p/7220369.html