How to Type
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5305 Accepted Submission(s): 2382
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 题解：
开一个二维数组，0代表当前输入结束lock是关着的；1代表开着的；
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x,y) scanf("%lf%lf",&x,&y)
#define P_ printf(" ")
const int MAXN=110;
typedef long long LL;
int dp[2][MAXN];
bool is_bigalpha(char c){
if(c>='A'&&c<='Z')return true;
else return false;
}
int main(){
int T;
char s[MAXN];
SI(T);
while(T--){
mem(dp,0);
scanf("%s",s+1);
int len=strlen(s+1);
dp[0][0]=0;
dp[1][0]=1;
for(int i=1;i<=len+1;i++){
if(is_bigalpha(s[i])){
dp[0][i]=min(dp[0][i-1]+2,dp[1][i-1]+2);
dp[1][i]=min(dp[0][i-1]+2,dp[1][i-1]+1);
}
else{
dp[0][i]=min(dp[0][i-1]+1,dp[1][i-1]+2);
dp[1][i]=min(dp[0][i-1]+2,dp[1][i-1]+2);
}
}
printf("%d
",min(dp[0][len],dp[1][len]+1));
}
return 0;
}