技術標籤：c++C語言c

There is a bookshelf which can fit n books. The i-th position of bookshelf is ai=1 if there is a book on this position and ai=0 otherwise. It is guaranteed that there is at least one book on the bookshelf.

In one move, you can choose some contiguous segment [l;r] consisting of books (i.e. for each i from l to r the condition ai=1 holds) and:

Shift it to the right by 1: move the book at index i to i+1 for all l≤i≤r. This move can be done only if r+1≤n and there is no book at the position r+1.
Shift it to the left by 1: move the book at index i to i−1 for all l≤i≤r. This move can be done only if l−1≥1 and there is no book at the position l−1.

For example, for a=[0,0,1,0,1] there is a gap between books (a4=0 when a3=1 and a5=1), for a=[1,1,0] there are no gaps between books and for a=[0,0,0] there are also no gaps between books.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤200) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1≤n≤50) — the number of places on a bookshelf. The second line of the test case contains n integers a1,a2,…,an (0≤ai≤1), where ai is 1 if there is a book at this position and 0 otherwise. It is guaranteed that there is at least one book on the bookshelf.

Output
For each test case, print one integer: the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without gaps).

Example
Input
5
7
0 0 1 0 1 0 1
3
1 0 0
5
1 1 0 0 1
6
1 0 0 0 0 1
5
1 1 0 1 1
Output
2
0
2
4
1
Note
In the first test case of the example, you can shift the segment [3;3] to the right and the segment [4;5] to the right. After all moves, the books form the contiguous segment [5;7]. So the answer is 2.

In the second test case of the example, you have nothing to do, all the books on the bookshelf form the contiguous segment already.

In the third test case of the example, you can shift the segment [5;5] to the left and then the segment [4;4] to the left again. After all moves, the books form the contiguous segment [1;3]. So the answer is 2.

In the fourth test case of the example, you can shift the segment [1;1] to the right, the segment [2;2] to the right, the segment [6;6] to the left and then the segment [5;5] to the left. After all moves, the books form the contiguous segment [3;4]. So the answer is 4.

In the fifth test case of the example, you can shift the segment [1;2] to the right. After all moves, the books form the contiguous segment [2;5]. So the answer is 1.

來源：
https://codeforces.com/problemset/problem/1433/B

大意：
就是填滿1之間的0，使之成為連續的1。求填入最小的1的個數。

思路：
包夾思想（數學上稱為夾逼定理）哈哈：
就是從開頭和結尾分別向裡拓展，遇見0就跳過，遇見1停下，計數，算兩個數的排列之間有多少個0。
具體看程式碼->
程式碼:

// C
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdalign>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif
using  namespace std;

int a[55]={0};

int main()
{
	int t;
	cin>>t;
	while(t--){
		int n,I,J,count=0;
		cin>>n;
		for(int i=1;i<=n;i++) cin>>a[i];
		I=1;J=n;
		while(a[I]==0){
			if(I==n) break;
			else I++;
		}
		while(a[J]==0){
			if(J==1) break;
			else J--;
		}
		if(I==n||J==1) cout<<0<<endl;
		else{
			for(int i=I;i<=J;i++){
				if(a[i]==0) count++;
			}
			cout<<count<<endl; 
		}
	}
	return 0;
}
``

//c++有點好用