有谁知道如何创建一个批处理文件,在一个目录中创建文件夹,增量名称为"问题1","问题2",...
我必须在两个给定的城市代码之间找到火车旅程,如果没有直接路线,那么我应该通过其他旅程找到间接路线.如果我想从A到B,我可能不得不从A到C到B.
我的火车路线文件格式为:出发码目的地代码公司价格时间这是两个城市代码之间的直接路线.
现在我使用以下循环进行直接连接,它可以工作,我只需要间接连接的帮助.
// load file data into v1
string dep, dest;
cout << "\n\tEnter the departure: ";
cin >> dep;
cout << "\n\tEnter the destination: ";
cin >> dest;
for(int i = 0; i < v1.size(); ++i) {
// Departure() and Destination(), return the departure/destination codes
if (v1[i].Departure() == dep && v1[i].Destination() == dest)
// here I find all the direct routes
else
// indirect routes dealt with here
}
我认为对于间接路线,我必须在其他部分处理它们.但我很难看到我会怎么做,我想我必须看看第一个出发地的目的地,并将其与我给定的目的地相匹配.
我正在制作一个简单的立方根猜测游戏,其中生成随机并显示它的立方体,然后用户输入立方根.这是我的计划:
int main()
try {
int max, min;
max = 99; min = 1; // only cubes of 1-99 are displayed
// display the title
cout << "\n\t\t\t\tCube Root Game" << endl;
cout << "\t\t\t\t=============\n" << endl;
srand(time(0)); // seed for random number generator
// display 10 numbers for the user to guess the cube root
for (int i = 0; i < 10; i++) {
int answer; // answer inputted by the user
int temp = rand() % (max …我有两个.php文件:
test1.php
<html>
<head>
<title>Sample</title>
</head>
<body>
<form action="test2.php" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
</body>
</html>
test2.php
<html>
<head>
<title>Sample Display Page</title>
</head>
<body>
<?php
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
?>
</body>
</html>
我想知道如何才能将它全部显示在单个页面上,即在test1.php上,而不必拥有两个文件.
我创建了以下类来对字符串数组进行排序.
public class StringSort {
private String[] hotelNames;
private int arrayLength;
public void sortHotel(String[] hotelArray) {
if (hotelArray.length <= 1) {
return;
}
this.hotelNames = hotelArray;
arrayLength = hotelArray.length;
quicksort(0, arrayLength - 1);
}
private void quicksort(int low, int high) {
int i = low, j = high;
String first = hotelNames[low];
String last = hotelNames[high];
String pivot = hotelNames[low + (high - low) / 2];
while( (first.compareTo(last)) < 0 ) { // first is less than last
while( (hotelNames[i].compareTo(pivot)) …