根据秒数获取要运行的循环

Question

我很难让循环在秒上运行int n = 1580;//number of moves;应该是秒数，但我得到的秒数输出不同。

我也很难让X和Y坐标正确，有时坐标正确，但有时不正确。

public class Steps {
public static void main(String[] args) {
int  forward1, left1;
forward1 = left1 = 1;

int forward2, left2, back, right;
left2 = back = 0;
forward2 = right = 1;
//counts for X coordinate
int countLeft = 0;//count moves left moves
int countRight = 0;//count moves right moves

//counts for Y coordinate
int countForward = 0;//count moves forward moves
int countBack = 0;//count moves right moves

int n =  11;//number of moves;
int i = 1;//iterator
int all = 0;//count total seconds

for(i=1; i<n;i++) {//the algorithm focuses on steps and not seconds.
if(i<n) {
System.out.println("Forward " + forward1);
left2 += 1;
i+=1;
all++;
//System.out.println(countForward);


if(i<n) {
for(int j = 0; j < left2; j++) {
System.out.println("left: " + left2);
countLeft++;
all++;//count seconds
}
back += 1;
i+=1;


//System.out.println(countLeft);

if(i<n) {
for(int j=0;j<back;j++) {
System.out.println("back: " + back);
countBack++;
all++;
//System.out.println(countBack);
}

left2 += 1;
i+=1;
back += 1;



if(i<n) {
System.out.println("left: "+left1);
forward2++;
i+=1;
all++;//count seconds
//System.out.println(countLeft);
countLeft++;
//System.out.println(countLeft);

if(i<n) {
for(int j = 0; j < forward2; j++) {
System.out.println("forward: " + forward2);
countForward++;
all++;
}
forward2++;
i+=1;
right +=1;

if(i<n) {
for(int j =0; j<right;j++) {
System.out.println("right: "+right);
countRight++;
all++;
}

right +=1;
i+=1;

}
}
}
} 
}   

}       
}

//System.out.println("Y"+((countForward )-countBack));
//in order to find the X co-ordinates we need to minus 
//the right moves from the left moves

//System.out.println(countLeft);
//System.out.println("right "+right);

////in order to find the Y co-ordinates we need to minus 
//the forward moves from the backward moves

//System.out.println("front " + countForward);
//System.out.println("back "+ countBack);

System.out.println("after " + (all) + " seconds");
System.out.println("Coordinate X: " + (countLeft - countRight));
System.out.println("Coordinate Y: " + (countForward - countBack ));

}}

当我用小数字运行代码时，时间似乎是正确的，当我用大数字运行n秒时，时间就不奇怪了。例如，当我运行程序n＝10时，我得到12秒。我得到以下输出：

Forward 1
left: 1
back: 1
left: 1
forward: 2
forward: 2
right: 2
right: 2
Forward 1
left: 3
left: 3
left: 3
after 12 seconds
Coordinate X: 3
Coordinate Y: 1

X坐标应为0，Y=3；

该算法的工作原理如下。

forward = 1
left = 1
back = 1
left = 1
forward = 2
right = 2
//end of loop
forward = 1
left = 3
back = 3
left = 1
forward = 4
right = 4
//end of loop
forward = 1
left = 5
back = 5
left = 1
forward = 6
right = 6

它会创建一个长方体。

一旦代码运行了一定的秒数，我需要能够提供它停止位置的X、Y坐标。

我最终使用调试器找到了问题，这个问题已经解决了，这要归功于

Answer 1

所以在运行调试器后，我发现我的X和Y坐标是正确的。然后，我在i变量上运行调试器。i注意到，当程序在三个for循环中的一个循环中迭代时，直到循环完成后，它才检查i变量是否大于n。这导致程序超过n，因此我们计算的秒数比n多。下面的代码解决了这个问题：

public class Steps {
public static void main(String[] args) {
int  forward1, left1;
forward1 = left1 = 1;

int forward2, left2, back, right;
left2 = back = 0;
forward2 = right = 1;
//counts for X coordinate
int countLeft = 0;//count moves left moves
int countRight = 0;//count moves right moves

//counts for Y coordinate
int countForward = 0;//count moves forward moves
int countBack = 0;//count moves right moves

//n = number of seconds this number will 
//have 1 added to it
int n =  43;
int i = 0;//iterator
int all = 0;//count total seconds

while(i<n) {//algorithm start

if(i<n) {
System.out.println("Forward " + forward1);
left2 += 1;
all++;
i++;
countForward++;
//System.out.println(countForward);


if(i<n) {
for(int j = 0; j < left2; j++) {
//added if to make sure loop ends 
//before i is larger than n
if(i<n) {
System.out.println("left: " + left2);
countLeft++;
all++;//count seconds
i++;    
}

}
back += 1;
//System.out.println(countLeft);


if(i<n) {
for(int j=0;j<back;j++) {
//added if to make sure loop ends 
//before i is larger than n
if(i<n) {
System.out.println("back: " + back);
countBack++;
all++;
i++;
//System.out.println(countBack);
}

}
left2 += 1;
back += 1;                  


if(i<n) {
System.out.println("left: "+left1);
forward2++;
all++;//count seconds
i++;
//System.out.println(countLeft);
countLeft++;
//System.out.println(countLeft);


if(i<n) {
for(int j = 0; j < forward2; j++) {
//added if to make sure loop ends 
//before i is larger than n
if(i<n) {
System.out.println("forward: " + forward2);
countForward++;
all++;
i++;
}
}
forward2++;
right +=1;


if(i<n) {
for(int j =0; j<right;j++) {
//added if to make sure loop ends 
//before i is larger than n
if(i<n) {
System.out.println("right: "+right);
countRight++;
all++;
i++;
}
}                                   
right +=1;

}
}
}
} 
}   

}       
}

//System.out.println("Y"+((countForward )-countBack));
//in order to find the X co-ordinates we need to minus 
//the right moves from the left moves

//System.out.println(countLeft);
//System.out.println("right "+right);

////in order to find the Y co-ordinates we need to minus 
//the forward moves from the backward moves

//System.out.println("front " + countForward);
//System.out.println("back "+ countBack);

System.out.println("after " + (all+1) + " seconds");
System.out.println("Coordinate X: " + (countLeft - countRight));
System.out.println("Coordinate Y: " + (countForward - countBack ));

}}

感谢那些通知我使用调试器的人。似乎当你在盒子里的时候，很难跳出盒子思考。

Answer 2

代码中的while循环当前正在运行，直到i大于或等于n。现在，如果要将n更改为程序应该运行的秒数，则需要将i更改为程序已经运行的秒。最有可能的是，无论你在什么平台上运行这个程序，都需要不到1850秒的时间来完成运行，所以当你运行它的时间更长时，你会多次迭代while循环，从而改变你得到的结果。