C/C++ programming learning-week 8⑦ skip class action

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Title description

Since going to university, Susu has learned to skip class=. =! However, he still has choices when he skips classes, and he has to finish listening to certain courses when he is kneeling. Now, given the course sequence of a certain day, use 1 to indicate an important course and 0 to indicate a less important course. Susu always skips class according to the following rules:

1. Never take important courses, never! ；
2. All the lessons before the first important lesson are knocked out;
3. If there are two or more non-important courses connected, all these non-important courses are knocked out;
4. Knock out all the lessons after the last important course;
5. If there are no important courses on the day, knock out all courses.
6. Under the guidance of such an advanced thought of skipping classes, how many classes can Susu take every day?

enter:

First enter a positive integer T (T<=150), which means the number of groups to be tested. Each of the following test data contains two rows. The first line is a positive integer n (1<=n<=1000), indicating the total number of courses on a certain day, and the second line has n integers (only 0 or 1), indicating the importance of these n courses.

Output:

For each set of test data, output the number of lessons Su Suneng took.

Sample Input

2
5
0 1 0 1 1
7
1 0 1 0 0 1 0

Sample Output

Case #1:4
Case #2:4

Ideas

Skip the class according to the skipping rules given by the title, do not skip the class with the number 1; skip all classes before the first 1 appears; if there are two or more 0s between 1 and 1, then these 0s All classes are skipped; all classes after the last 1 class are skipped; if there is no class 1, all classes of the day are skipped.

C++ code:

#include<bits/stdc++.h>
using namespace std;
int n, t, num = 0, first, last, a[1005];
int main()
{
    
    
	
	while(cin >> t)
	{
    
    
		for(int i = 0; i < t; i++)
		{
    
    
			cin >> n;
			num = 0, first = -1, last = -1;
			memset(a, 0, sizeof(a));
			for(int j = 0; j < n; j++)
			{
    
    
				cin >> a[j];
				if(a[j] == 1 && first == -1)
				{
    
    
					num++;
					first = j;
					last = j;
				}
				else if(a[j] == 1)
				{
    
    
					num++;
					last = j;
				}
			}
			for(int j = first; j < last; j++)
				if(a[j] == 0&&a[j - 1] != 0 && a[j + 1] != 0)
					num++;
			cout << "Case #" << i + 1 << ":" << num << endl;
		}
	}
	return 0;
}