CF--1649A. Game

最新推荐文章于 2025-12-03 21:54:37 发布

原创最新推荐文章于 2025-12-03 21:54:37 发布 · 235 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#p2p #linq #蓝桥杯

本文介绍了一个关于计算机游戏中的关卡设计问题，玩家需要从第一个到最后一个地点，中间可能包含陆地和水域。玩家只能在陆地上移动，并且可以进行跳跃，但需要消耗金币。每个跳跃可以跳过最多x个位置。任务是找出从起点到终点所需的最低金币数。通过示例解释了如何计算最短路径，并提供了解决方案的代码实现。

You are playing a very popular computer game. The next level consists of nn consecutive locations, numbered from 11 to nn, each of them containing either land or water. It is known that the first and last locations contain land, and for completing the level you have to move from the first location to the last. Also, if you become inside a location with water, you will die, so you can only move between locations with land.

You can jump between adjacent locations for free, as well as no more than once jump from any location with land ii to any location with land i + xi+x, spending xx coins (x \geq 0x≥0).

Your task is to spend the minimum possible number of coins to move from the first location to the last one.

Note that this is always possible since both the first and last locations are the land locations.

Input

There are several test cases in the input data. The first line contains a single integer tt (1 \leq t \leq 1001≤t≤100) — the number of test cases. This is followed by the test cases description.

The first line of each test case contains one integer nn (2 \leq n \leq 1002≤n≤100) — the number of locations.

The second line of the test case contains a sequence of integers a_1, a_2, \ldots, a_na1,a2,…,an (0 \leq a_i \leq 10≤ai≤1), where a_i = 1ai=1 means that the ii-th location is the location with land, and a_i = 0ai=0 means that the ii-th location is the location with water. It is guaranteed that a_1 = 1a1=1 and a_n = 1an=1.

Output

For each test case print a single integer — the answer to the problem.

Sample 1

Inputcopy	Outputcopy
3 2 1 1 5 1 0 1 0 1 4 1 0 1 1	0 4 2

Note

In the first test case, it is enough to make one free jump from the first location to the second one, which is also the last one, so the answer is 00.

In the second test case, the only way to move from the first location to the last one is to jump between them, which will cost 44 coins.

In the third test case, you can jump from the first location to the third for 22 coins, and then jump to the fourth location for free, so the answer is 22. It can be shown that this is the optimal way.

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n;
		cin>>n;
		int a[110];
		int f1=0,f2=0;
		for(int i=1;i<=n;i++)
		{
			cin>>a[i];
			if(a[i]==0&&a[i-1]==1&&!f1)
			{
				f1=i;
			}
			else if(a[i]==1&&a[i-1]==0)
			{
				f2=i;
			}
		}
		if(f1)
		{
		//	cout<<f2<<" "<<f1<<endl;
			cout<<f2-f1+1<<endl;
		}
		else
		cout<<0<<endl;
		
	
		
	}
	return 0;
}