Codeforces Round 56-A. Dice Rolling(思维题)

本文探讨了一个关于Mishka使用特殊六面骰子(2到7的整数)来达到特定点数的问题。通过分析,我们发现只要将目标点数除以2,即可得出Mishka至少需要掷骰的次数,从而达到指定点数。

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time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mishka got a six-faced dice. It has integer numbers from 22 to 77 written on its faces (all numbers on faces are different, so this is an almostusual dice).

Mishka wants to get exactly xx points by rolling his dice. The number of points is just a sum of numbers written at the topmost face of the dice for all the rolls Mishka makes.

Mishka doesn't really care about the number of rolls, so he just wants to know any number of rolls he can make to be able to get exactly xxpoints for them. Mishka is very lucky, so if the probability to get xx points with chosen number of rolls is non-zero, he will be able to roll the dice in such a way. Your task is to print this number. It is guaranteed that at least one answer exists.

Mishka is also very curious about different number of points to score so you have to answer tt independent queries.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of queries.

Each of the next tt lines contains one integer each. The ii-th line contains one integer xixi (2≤xi≤1002≤xi≤100) — the number of points Mishka wants to get.

Output

Print tt lines. In the ii-th line print the answer to the ii-th query (i.e. any number of rolls Mishka can make to be able to get exactly xixi points for them). It is guaranteed that at least one answer exists.

Example

input

Copy

4
2
13
37
100

output

Copy

1
3
8
27

Note

In the first query Mishka can roll a dice once and get 22 points.

In the second query Mishka can roll a dice 33 times and get points 55, 55 and 33 (for example).

In the third query Mishka can roll a dice 88 times and get 55 points 77 times and 22 points with the remaining roll.

In the fourth query Mishka can roll a dice 2727 times and get 22 points 1111 times, 33 points 66 times and 66 points 1010 times.

每次都是两道,唉,何时才能提高啊

题解:把每次都看成2即可,感觉有点水

代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;


int main()
{
	
	int n;
	cin>>n;
	int k;
	for(int t=0;t<n;t++)
    {
    	cin>>k;
    	cout<<k/2<<endl;
	 } 
	
	return 0;
} 

 

转载于:https://www.cnblogs.com/Staceyacm/p/10781922.html

根据提供的引用内容,Codeforces Round 511 (Div. 1)是一个比赛的名称。然而,引用内容中没有提供这个比赛相关的具体信息或问。因此,我无法回答关于Codeforces Round 511 (Div. 1)的问。如果您有关于这个比赛的具体问,请提供更多的信息,我将尽力回答。 #### 引用[.reference_title] - *1* [Codeforces Round 860 (Div. 2)解](https://blog.csdn.net/qq_60653991/article/details/129802687)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v91^control,239^v3^insert_chatgpt"}} ] [.reference_item] - *2* [Codeforces Round 867 (Div. 3)(A到E)](https://blog.csdn.net/wdgkd/article/details/130370975)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v91^control,239^v3^insert_chatgpt"}} ] [.reference_item] - *3* [Codeforces Round 872 (Div. 2)(前三道](https://blog.csdn.net/qq_68286180/article/details/130570952)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v91^control,239^v3^insert_chatgpt"}} ] [.reference_item] [ .reference_list ]
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