2 回答

TA貢獻1825條經(jīng)驗 獲得超6個贊
我在為第二部分輸入時感到震驚,這將包括一個 M 時間循環(huán),每次輸入 4 個輸入,分別為 IJKL
根據(jù)您的要求,我建議您創(chuàng)建一個類而不是使用原始類型。這比制作二維數(shù)組更容易,而且您可以對數(shù)據(jù)做更多的事情。
由于我不完全知道這個問題,我會假設(shè)(建筑物的地板、建筑物的窗戶、爬上建筑物所花費的時間、過馬路所花費的時間)都是道路的屬性。如果沒有,那么您可以對建筑物進行同樣的操作并將屬性從道路移動到建筑物
道路類
public class Road {
private int floorsOfBuilding;
private int windowsInBuilding;
private int climbTime;
private int crossingTime;
// Constructor
public Road(int floorsOfBuilding, int windowsInBuilding, int climbTime, int crossingTime) {
this.floorsOfBuilding = floorsOfBuilding;
this.windowsInBuilding = windowsInBuilding;
this.climbTime = climbTime;
this.crossingTime = crossingTime;
// You can do simple calculations here but for more complex it is better
// to create a method to maintain readability
}
public String toString() {
return "\n" + floorsOfBuilding + " " + windowsInBuilding + " " + climbTime + " " + crossingTime;
}
}
主課
int roadsCount = Integer.parseInt(buildingsAndRoadsArray[1]);
Road[] roadsArray = new Road[roadsCount];
for (int i = 0; i < roadsArray.length; i++) {
System.out.println();
System.out.println("Road:" + i);
System.out.print("Enter I J K L: ");
String input = scan.nextLine();
String[] inputSplit = input.split(" ");
Road road = new Road(Integer.parseInt(inputSplit[0]), Integer.parseInt(inputSplit[1]),
Integer.parseInt(inputSplit[2]), Integer.parseInt(inputSplit[3]));
roadsArray[i] = road;
}
scan.close();
System.out.println();
for (int i = 0; i < roadsArray.length; i++) {
System.out.println("Road: " + i + roadsArray[i] + "\n");
}
輸出
Enter Number of Buildings and Roads: 1 2
// Input
Road:0
Enter I J K L: 12 13 14 15
Road:1
Enter I J K L: 49 59 69 79
// Output
Road: 0
12 13 14 15
Road: 1
49 59 69 79

TA貢獻1946條經(jīng)驗 獲得超3個贊
for (int i = 0; i < M; i++) {
int I = sc.nextInt();
int J = sc.nextInt();
int K = sc.nextInt():
int L = sc.nextInt();
// your code here
}
由于輸入是常數(shù),您可以直接輸入數(shù)字。
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