How to Type（字符大小写转换）

How to Type（字符大小写转换）

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 8
The string “HDUACM”, can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8
代码

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    int i,j,n,len;    scanf("%d",&n);    getchar();    while(n--)    {        char a[105];        int sum=0,x=0;        gets(a);        len=strlen(a);        for(i=0;i<len;)        {            if(x==0&&a[i]>='A'&&a[i]<='Z'&&a[i+1]>='A'&&a[i+1]<='Z')//如果连续两个大写 就转成大写的；            {                x=1;                sum++;            }            if(x==1&&a[i]>='a'&&a[i]<='z'&&(a[i+1]>='a'&&a[i+1]<='z'||a[i+1]==0))////连续两个小写的就转成小写的                        //注意：如果已经是最后一个的话，就要转换成小写，这样最后可以少一步转换回小写的操作；            {                x=0;                sum++;            }            if(a[i]>='a'&&a[i]<='z')            {                if(x==0)                {                    sum++;                    i++;                }                else                {                    sum+=2;                    i++;                }            }            else            {                if(x==0)                {                    sum+=2;                    i++;                }                else                {                    sum++;                    i++;                }            }        }        if(x==1)            sum++;        printf("%d\n",sum);    }    return 0;}