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;
}

最新文章

  1. [LeetCode] Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树
  2. 有用的css片段
  3. Sublime Text 3 中文乱码的解决方法
  4. [Android]竖直滑动选择器WheelView的实现
  5. ubuntu 编译安装 srilm
  6. linux rwxrwxrwt文件夹属性
  7. gcj_2016_Round1_B
  8. paip.提升用户体验----gcc c++ JIT-debugging 技术
  9. 【17】以独立语句将newed对象置入智能指针
  10. STL之set、multiset、functor&amp;pair使用方法
  11. 继承PictureBox显示GIF的自定义控件实现
  12. php fpm start.sh
  13. Week16(12月23日):复习
  14. 关于volatile的可见性和禁止指令重排序的疑惑
  15. html5,js插件实现手机端实现头像剪切上传
  16. [Luogu3936]Coloring
  17. Linux中一个快速查找文件和目录的命令
  18. 并查集(POJ1182)
  19. JavaWeb基础-Jsp的请求与响应
  20. 常用的Markdown语法

热门文章

  1. marlab之rangesearch()函数
  2. Struts2 小例子 --第二弹
  3. nginx的fastcgi_param参数详解
  4. sangfor-AF 地址转换以及各种模式理解(路由,透明,虚拟网线,混合模式)
  5. C++正确的cin输入
  6. python script
  7. 批处理中格式化Date
  8. ZigBee自组网地址分配与路由协议概述
  9. lwip【5】 lwIP配置文件opt.h和lwipopts.h初步分析之二
  10. java线程基本原理