## FOJ 1906解题报告

### Problem Description

There is a sequence that only consists of characters ‘/’ and ‘’. You should calculate the number of down sequences and up sequences respectively in the given sequence.
A down sequence is defined as follows:
1. “/” is a down sequence.
2. If S is a down sequence, then “S/” is also a down sequence.
3. Any other sequences are not a down sequences.
Similarly, an up sequence is defined as follows:
1. “/” is an up sequence.
2. If S is an up sequence, then “/S” is also an up sequence.
3. Any other sequences are not an up sequences.
For example, sequences “/”, “//” and “///” are all up sequences, while “/”, “/” and “//” are not. Sequences “/”, “//” and “///” are all down sequences, while “/”, “//” and “//” are not. There is only one down sequence in the sequence “//”, that is “/”. There are two up sequences in the sequence “//”, they are “/” and “//”.

### Input

The first line of the input contains an integer T (T <= 10), indicating the number of cases. Each case begins with a line containing one integer n (1n100), the length of the sequence. The next line contains the sequence, consisting of characters ‘/’ and ‘’.

### Output

For each test case, print a line containing the test case number (beginning with 1) and two numbers separated by a space indicating the number of down sequences and up sequences respectively in the given sequence.

3
2
/
7
////
15
//////////

Case 1: 0 1
Case 2: 1 3
Case 3: 5 5

### Source

2010年全国大学生程序设计邀请赛（福州）热身赛

### Code

```#include
#include
int main()
{
int t,n,x,i,count=0,up,down,dr,dl;
char ch[105];
bool flag;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
scanf("%s",ch);
up=0;
for(i=0;i
```