#include <iostream>
using namespace std;
int main()
{
    int a , b , c;
    cin >> a;
    b = a / 8;
    c = a % 8;
    if (c != 0)
    {
        b += 1;
    }
    cout << b;
    return 0;
}

{"messages":{},"UiContext":{"cdn_prefix":"/","url_prefix":"/","ws_prefix":"/","nav_logo_dark":"http://oj.hetao101.com/d/cpp_teacher/file/106/55.png","constantVersion":"e6b2fad1","domainId":"training","domain":{"_id":"training","lower":"training","owner":2,"name":"练习题库","bulletin":"# 核OJ_核桃编程 - 练习题库\r\n\r\n[⇐ 返回根题库](http://oj.hetao101.com/)\r\n\r\n## 基本使用方法\r\n\r\nP 系列题目来自网络，可能会有题目排版数据问题，部分纯输出的题目可能缺少数据，管理员看到缺少数据系统错误的提交后会及时修复。遇到问题可以在讨论区发布帖子。\r\n\r\n1. 通过题库点击某道想做的题目\r\n2. 读题\r\n3. 点击进入在线编程模式\r\n4. 点击右上方的自测打开自测窗口\r\n5. 将题目的样例输入复制进输入框中\r\n6. 点击运行自测，查看是否与输出一致\r\n7. 如果结果没有问题，可以点击递交评测查看评测结果\r\n8. 如果显示错误、可以对应地修改代码再次自测\r\n\r\n![](http://oj.hetao101.com/file/2/t.png)\r\n\r\n## 常见设置\r\n\r\n0. 设置当前题库的显示名：<http://oj.hetao101.com/d/training/home/settings/domain>\r\n1. 进入当前题库的偏好设置页面：<http://oj.hetao101.com/d/training/home/settings/preference>\r\n2. 将显示偏好中的**用户界面语言**设置成**简体中文**\r\n3. 将使用偏好中的**代码语言**设置成自己常用的语言\r\n4. 将使用偏好中的**默认代码模板**改成自己习惯的代码框架，比如下面这段\r\n\r\n```cpp\r\n#include<bits/stdc++.h>\r\nusing namespace std;\r\nint main(){\r\n\r\n\r\n\r\n    return 0;\r\n}\r\n```\r\n","roles":{"3":"10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376","teacher":"10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652777725396788478736373","guest":"10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624967874125500547201","default":"10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652777525100954235143873","root":"-1","cpp":"10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652789254612192640104439","contest_creator":"10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652777526956929862828997","new_teacher":"10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652777543124148872280005"},"avatar":"url:https://ecmb.bdimg.com/tam-ogel/1969563307_-1467192127_88_88.jpg","langs":"cc.std98,cc.std11,cc.std14,py.py3,cc,py","share":"*"},"SWConfig":{"preload":"","hosts":["http://oj.hetao101.com","https://oj.hetao101.com","oj.hetao101.com/","/"],"domains":[]}},"UserContext":"{\"_id\":17475,\"mail\":\"5976915@hetao101.com\",\"uname\":\"hetao5976915\",\"hashType\":\"hydro\",\"priv\":16842756,\"regat\":\"2023-07-12T06:13:59.355Z\",\"loginat\":\"2023-08-01T11:27:44.933Z\",\"perm\":\"BigInt::10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652777525100954235143873\",\"scope\":\"BigInt::-1\",\"role\":\"default\",\"domains\":[],\"tfa\":false,\"authn\":false,\"group\":[\"17475\"],\"timeZone\":\"Asia/Shanghai\",\"codeLang\":\"cc\",\"avatar\":\"gravatar:5976915@hetao101.com\",\"gender\":2,\"backgroundImage\":\"/components/profile/backgrounds/1.jpg\",\"unreadMsg\":0,\"pinnedDomains\":[],\"fontFamily\":\"Open Sans\",\"codeFontFamily\":\"Source Code Pro\",\"preferredEditorType\":\"sv\",\"monacoTheme\":\"vs-light\",\"theme\":\"light\",\"formatCode\":true,\"nAccept\":9,\"nSubmit\":23,\"rp\":5,\"rpInfo\":{\"problem\":5},\"rank\":5740,\"level\":1,\"avatarUrl\":\"//cn.gravatar.com/avatar/dc51bdcf0c080f7680e4d1069ddde2a2?d=mm&s=128\"}"}
那个大神看看😕 👎 

😄

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👎

一直以为最后一道题会很难，不料竟是送分的！

上代码

#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
if(n%8==o)  c0ut<<n/8;
else  c0ut<<n/8+1;
return o;
}

这条代码一点陷阱都没有🎉️

#include <iostream>
using namespace std;
int main()
{
    int a;//定义一个存放柴总数的整数变量a。
    cin>>a;//输入柴的总数。
    if (a%8==0)//判断a是否是8的倍数
    {
        cout<<a/8;
    }
    else//如果a不是8的倍数。
    {
        cout<<a/8+1;
    }
    return 0;
}

#include <iostream>
using namespace std;
int main()
{
    int a,b;
    cin>>a;
    if (a%8==0)
        b=a/8;
    else
        b=a/8+1;
    cout<<b;
}

#include <iostream>
using namespace std;
int main()
{
double n;
cin >> n;
//核心代码：
int a = n / 8 + 0.9;
cout << a;
return 0;
}
这样简单

刚刚是错的，这次就对了

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >>n;
    //定义+输入
    if (n%8==0)
    {
        cout <<n/8;
    }
    else
    {
        cout <<n/8+1;
    }
    // 判断
    return 0;
}

新手上路，请多关照！！

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a;
    cin >> a;
    int b = a / 8;
    if (a % 8 != 0) b ++;
    cout << b << endl;
	re turn 0; 
}

LTC LEVEL1中唯一的HT开头的题目 也是唯四贼简单的

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    if(n%8==0)
    {
        cout<<n/8;
    }
    else
    {
        cout<<n/8+1;
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    n % 8?
    cout << n/8 + 1:
    cout << n/8;
    return 0;
}

#include <iostream> using namespace std; int main() { int n; cin>>n; if (n%8==0) { cout<<n/8; } else { cout<<n/8+1; } }

太简单了

int a;
    cin >> a;
    if (a%8==0)
    {
        cout << a/8;
    }
    else
    {
        cout << a/8+1;
    }

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    if (n % 8 == 0)
    {
        cout << n / 8 << endl;
    }
    else
    {
        cout << n / 8 + 1 << endl;
    }
    return 0;
}

#include <iostream>
using namespace std;
int main()
{ 
    int n;
    cin >> n;
    if (n % 8 == 0)
    {
        cout << n/8;
    }
    else
    {
        cout << n/8+1;
    }
    return 0;
}

EASY

#include <iostream>
using namespace std;
int main () {
    int n, sum = 0; //定义
    cin >> n; //输入
    for (int i = n; i > 0; i -= 8) { //循环，每次循环变量都-8
        sum++; //次数
    }
    cout << sum; //输出
    return 0; //结束
}

#include <iostream>
using namespace std;
int main()
{
    long long n;//定义
    cin >>n;//输入
    long long sum=n/8;//定义
    if (n%8!=0)//判断是否有剩余
    {
        sum++;//如果有，那么再运一次
    }
    cout <<sum;//输出
    return 0;
}

#include <iostream>
using namespace std;

int main() {
    int num;
    cin >> num;

    int trips = num / 8;
    if (num % 8 != 0) {
        trips++;
    }

    cout << trips << endl;

    return 0;
}

在上述代码中，通过整除和取模运算计算出搬运的次数。将木柴的数量除以 8，得到的结果表示完整的搬运次数，然后使用判断是否有剩余的木柴需要额外搬运。如果有剩余，则搬运次数需要加 1。 记得点赞。

#include <iostream>
using namespace std;
int main()
{
    int a, b;
    cin >> a;
    b = a / 8;
    if (a % 8 != 0)
    {
        b += 1;
    }
    cout << b;
    return 0;
}