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.

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.

For each test case, you must output the smallest times of typing the key to finish typing this string.

Dellenge

类似于上次写的那道可以上下右走方格的题目，也是常见的一类DP，有一种分层的思想!

对于此题就在于Caps键在DP的过程中可能处于开/关两种，我们不妨多开一维0表示Caps键处于关闭1表示处于开启状态.

直接地推1Ahhh,还是在喝了酒之后好开森>_<

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
int dp[105][2];
char s[105];
int solve()
{
int i,j,k,sz=strlen(s);
memset(dp,inf,sizeof(dp));
if(islower(s[0])){
dp[0][0]=1;
dp[0][1]=2;
}
else{
dp[0][0]=2;
dp[0][1]=2;
}
for(i=1;i<sz;++i){
if(islower(s[i])){
dp[i][0]=min(dp[i-1][1]+2,dp[i-1][0]+1);
dp[i][1]=min(dp[i-1][1]+2,dp[i-1][0]+2);
}
else{ //da xie
dp[i][0]=min(dp[i-1][1]+2,dp[i-1][0]+2);
dp[i][1]=min(dp[i-1][1]+1,dp[i-1][0]+2);
}
}
return min(dp[sz-1][0],dp[sz-1][1]+1);
}
int main()
{
int t,n,m,i,j;
cin>>t;
while(t--){cin>>s;
cout<<solve()<<endl;
}
return 0;
}