CF1649A.Game
普及/提高-
通过率:0%
AC君温馨提醒
该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。
题目描述
You are playing a very popular computer game. The next level consists of n consecutive locations, numbered from 1 to n , 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 i to any location with land i+x , spending x coins ( x≥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.
输入格式
There are several test cases in the input data. The first line contains a single integer t ( 1≤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 n ( 2≤n≤100 ) — the number of locations.
The second line of the test case contains a sequence of integers a1,a2,…,an ( 0≤ai≤1 ), where ai=1 means that the i -th location is the location with land, and ai=0 means that the i -th location is the location with water. It is guaranteed that a1=1 and an=1 .
输出格式
For each test case print a single integer — the answer to the problem.
输入输出样例
输入#1
3 2 1 1 5 1 0 1 0 1 4 1 0 1 1
输出#1
0 4 2
说明/提示
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 0 .
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 4 coins.
In the third test case, you can jump from the first location to the third for 2 coins, and then jump to the fourth location for free, so the answer is 2 . It can be shown that this is the optimal way.