B. Yet Another Bookshelf
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
Your task is to find the minimum number of moves required to collect all the books on the shelf as a contiguous (consecutive) segment (i.e. the segment without any gaps).
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.
思路：
把 连续的1 只看作 一个1
然后找出第一个1和最后一个1之间有几个0就可以
代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+50;
int main()
{
int t,n,x;
cin>>t;
while(t--)
{
cin>>n;
vector<int> v;
int last=-1,f,r; //last记录上一个数，f第一个1，r最后一个1
for(int i=0;i<n;i++)
{
cin>>x;
if(x && x!=last) v.push_back(x); //不重复才加入
else if(x==0) v.push_back(0);
last=x;
}
for(int i=0;i<v.size();i++) //找到第一个1的位置
{
if(v[i]==1)
{
f=i;
break;
}
}
for(int i=v.size()-1;i>=0;i--) //找到最后一个1的位置
{
if(v[i]==1)
{
r=i;
break;
}
}
int ans=0;
for(int i=f;i<=r;i++) //这个区间内0的个数就是答案
{
if(v[i]==0) ans++;
}
cout<<ans<<endl;
}
return 0;
}