Максим Филимонов's questions

输入 N - 矩阵的大小，K - 子矩阵的大小。
接下来引入一个矩阵
Let N = 3, K = 2

我们的 N*N 矩阵

1 2 1
1 1 2
1 1 1

所有大小为 K*K 的子矩阵：

[1 2
 1 1]

[2 1
 1 2]

[1 1
 1 1]

[1 2
 1 1]

如何对子矩阵进行相同的发现？从什么推回来？他唯一做的就是创建一个由 k*k 矩阵组成的三维数组，其中填充了零。

var n, c, i, x: integer;
begin 
 readln(n);
 c := 0;
 for i := 1 to n do
    readln(x);
    if ((x >= 100) and (x <= 999) and (x div 100 = 3) and (x mod 5 = 0)) then c := c + 1;
 writeln(c);
end.

它必须从长度为 n 的序列中找到以 3 开头且可被 5 整除的三位数字的个数。

#include <iostream>
#include <algorithm>

using namespace std;

struct vector {
    vector *nextElem;
    int ans;
    int length;
    vector(int n) {
        this->ans = 0;
        vector *next = nextElem;
        for (int i=0; i<=n; i++){
            next->ans=0;
            next=next->nextElem;
        }
        length = n;
     }
     vector(int n, int val) {
        this->ans = val;
        vector *next = nextElem;
        for (int i=0; i<=n; i++){
            next->ans=val;
            next=next->nextElem;
         }
         length = n;
     }
     void push_back(int val) {
         vector *next = nextElem;
         for (int i = 1; i < length; i++) {
               next=next->nextElem;
         }
         length++;
         next->ans = val;
     }
     void pop_back() {
         vector *next = nextElem;
         for (int i = 0; i < length; i++) {
               next=next->nextElem;
         }
         length--;
         next->nextElem=0;
     }
    int get(int i) {
       vector *next = nextElem;
       if (i > 0) {
            for (int j = 1; j < i; j++) {
                   next=next->nextElem;
            }
        return next->ans;
       } else return ans;
    }
    int size() {
        return length;
    }
};





int main() {
    vector *v;
    v = new vector(2, 10);
    v->push_back(0);
    std::cout << v->size() << std::endl; // 3
    std::cout << v->get(2) << std::endl; // 0
    std::cout << v->get(1) << std::endl; // 10
    std::cout << std::endl;
    v->pop_back();
    v->pop_back();
    std::cout << v->size() << std::endl; // 1
    std::cout << std::endl;
    v->push_back(5);
    v->push_back(17);
    v->push_back(3);
    std::cout << v->get(0) << std::endl; // 10
    std::cout << v->get(1) << std::endl; // 5
    std::cout << v->get(3) << std::endl; // 3
    std::cout << v->size() << std::endl; // 4
}

带一个参数的构造函数会创建一个包含 n 个元素的堆栈，其值为 0；有两个参数 - 值为 val。push_back 和 pop_back 方法在堆栈上创建和删除元素。get方法通过i参数返回元素的值，size也返回序列中元素的个数。项目计数从 0 开始。

一切似乎都运行良好，但执行后，Code::Blocks 显示消息Process returned -1073741571 (0xC00000FD) execution time : 1.473 s。帮助

struct my_time {
    long long day, hour, minute, second;
    void set(long long t) {
        if (t>=0) {
            day = t/60/60/24;   
            hour = t/60/60-day*24;
            minute = t/60-hour*60-day*24*60;
            second = t-minute*60-hour*60*60-day*60*60*24;
        } else if (t%86400==0) {
            day=t/86400;
            hour=0;
            minute=0;
            second=0;
        } else  {
            day = -1;
            day += t/86400;
            hour = abs(t/60/60-24*day-1);
            minute = abs(t/60-hour*60-24*60*day-1);
            second = (abs(t-minute*60-hour*60*60-60*60*24*day-1)+1)%60;
        }
    }
};

需要写一个时间结构体，set方法接收以秒为单位的秒数作为参数。当我进入方法时出现错误，例如-600。这些字段等于-1 23 49 0，但它们应该是-1 23 50 0。告诉我问题出在哪里。

#include <iostream>

using namespace std;

int main() {
    int n;
    cin >> n;
    long long arr1[n];
    for (int i = 0; i < n; i++) {
        cin >> arr1[i];
    }
    for (int i = 0; i < n; i++) {
        for (int k = i; k < n; k++) {
            if (arr1[k] % 2 == 0) {
                long long t = arr1[k];
                arr1[k] = arr1[i];
                arr1[i] = t;
            }
        }
        cout << arr1[i] << " ";
    }
}

有必要重新排列数组中的偶数和奇数，使偶数在开头，奇数仅在它们之后。必须保持顺序

数字已重新排列，但顺序未保留，请帮助

如何将长度设置<div>为浏览器窗口长度的四分之一？

var x, x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11: integer;
begin
read(x);
x1 := trunc(x);
x2 := x1 div 100;
x3 := x1 div 10 mod 10;
x4 := x1 mod 10;
x5 := x2 + x3 + x4;
x6 := x - x1;
x7 := x6 div 1000;
x8 := x6 div 100 mod 10;
x9 := x6 div 10 mod 10;
x10 := x6 mod 10;
x11 := x7 + x8 + x9 + x10;
writeln(x5);
writeln(x11);
end.

使用时与<input>图片一起移动margin。怎么才能动<input>？

#fullnameinput {
  height: 46px;
  width: 281px;
  background-color: rgba(255, 255, 255, 1);
  border-radius: 2px 2px 2px 2px;
  font-size: 13px;
  background-image: url(48.png);
  background-repeat: no-repeat;
  background-position: 4px;
  outline: none;
  -moz-border-radius: 5px;
  -webkit-border-radius: 5px;
  padding-left: 20px;
  margin-top: 289px;
  margin-left: 306px;
  border: none;
  position: relative;
}

<header>
  <div id="logo1"><img src="888.png"></div>
  <nav>
    <div id="menu">
      <a class="menu__item" href="">HOME</a>
      <a class="menu__item" href="">SERVICE</a>
      <a class="menu__item" href="">ABOUT US</a>
      <a class="menu__item" href="">PRICING TABLE</a>
      <a class="menu__item" href="">HOW IT WORK</a>
      <a class="menu__item" href="">HAPPY CLIENTS</a>
      <a class="menu__item" href="">CONTACT US</a>
    </div>
  </nav>
</header>
<div id="banner12345">
  <p id="mainheader"><strong>BOOST UP YOUR LOCAL BUSINESS</strong></p>
  <p id="maintext">Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum typesetting.</p>
  <input placeholder="    Full Name" id="fullnameinput" type="text" name="Full Name">

  <img id="bannerpng" width="100%" src="20.png">

</div>