学生储物柜拼图

Question

/* 学生100人，储物柜100个。学生 1 全部打开，学生 2 每秒钟关闭一次，学生 3 每三分之一更改一次储物柜（如果打开则关闭，如果关闭则打开），学生 4 每隔四个储物柜更换一次，依此类推所有 100 名学生。哪个储物柜将保持打开状态？ */

到目前为止，这是我的代码：

#include <stdio.h>

int main(void)
{
int locker[100], i, closed = 0, opened = 0;

for(i=0; i<100; i++) locker[i] = 1;//0 means closed locker, 1 means open locker
for(i=1; i<101; i++) if(i % 2 == 0) locker[i-1] = 0; // every second locker is closed by second student...(2,4,6,7)
for(i=3; i<101; i++){ // i means student no. i
    if(locker[i-1] == 0) locker[i-1] = 1;
    if(locker[i-1] == 1) locker[i-1] = 0;

如果我用 "else" 替换 "if(locker[i-1] == 1)" 为什么程序不起作用？正确的结果是opened 1 closed 99.如果我用'else'结果变成opened 50 and closed 50

}
for(i=0; i<100; i++){
    if(locker[i] == 0) closed = closed + 1;
    else opened = opened + 1;
}
printf("opened locker %d\nclosed locker %d", opened, closed);
return 0;

}

这是我在堆栈溢出中的第一个 post。如果我做错了什么，请纠正我。

Answer 1

虽然我没有检查完整的代码，而且我对你的问题的逻辑不是很清楚，但这些行在你的代码中似乎是有问题的：

if(locker[i-1] == 0) locker[i-1] = 1;
if(locker[i-1] == 1) locker[i-1] = 0;

你在这里做的是，如果一个值是 0，那么你将它设置为 1，然后你再次检查，如果它是 1，你将它设置为 0。所以，所以在这种情况下通过这两个语句运行之后，所有值都将设置为 0。

相反，你应该这样做

if(locker[i-1] == 0) locker[i-1] = 1;
else locker[i-1] = 0;

Answer 2

请注意，您的循环是错误的，因为您正在循环第三个学生的每个储物柜，而不是循环剩余的储物柜。您应该为每个学生 (n) 将第 n 个储物柜换成相反的。

当你连续有两个 if 时。如果第一个 if 打开储物柜，第二个 if 看到它打开并关闭它（这是错误的）。需要 else 才能实际更改它。

还有一点就是可以用exclusive or代替if locker[i] ^= 1

#include <stdio.h>

int main(void)
{
int locker[100], i, k, closed = 0, opened = 0;

for(i=0; i<100; i++) {
  if (i%2 == 0) locker[i] = 1; // odd lockers (base 1) stay open
  else locker[i] = 0; // even lockers (base 1) are closed
  //0 means closed locker, 1 means open locker
  for(i=3; i<101; i++){ // i means student no. i
    for (k=i; k<101); k+=i) { // change every ith locker
      // if (locker[k-1] == 0) locker[k-1]=1
      // else locker[i-1] = 0;
      // use exclusive or instead of if
      locker[i-1] ^= 1;
    }
   }
}
// Now check the number open or closed
for(i=0; i<100; i++){
    if(locker[i] == 0) closed = closed + 1;
    else opened = opened + 1;
}
printf("opened locker %d\nclosed locker %d", opened, closed);
return 0;

}

Answer 3

我会给你一些提示来帮助你。

答案是 10 个储物柜保持打开状态，90 个关闭。
对于这个特定问题，如果避免从零开始的索引，编写代码会更容易。所以将数组声明为 int locker[101]; 然后使用索引 1 到 100代表100个储物柜。
第 N 个学生应该更换第 N 个储物柜。所以你需要两个嵌套的 for 循环。外循环跟踪 n，而内循环翻转储物柜。
只影响第 N 个储物柜的内部循环应该是这样的这个
```
for ( i = n; i <= 100; i += n ) // every Nth locker
    locker[i] = 1 - locker[i];  // flip the locker
```
请注意，我们有 i=n 而不是正常的 i=0 和 i++ 和 i+=n。因此，例如，如果 n 为 3，则 i 的值是 3,6,9,...

Answer 4

LOCKERS 是新的 FIZZBUZZ。

您已经对这个问题有了很好的答案。这是一个使用布尔值的小程序。

 /* lockers.c

   There are 100 students and 100 lockers. Student 1 opens all, student 2
   closes every second one, student 3 changes every third locker (closes if
   open, opens if close), Student 4 changes every forth locker and so on for
   all 100 students. Which lockers will be open at the end of the process?
*/

#include <stdio.h>
#include <stdbool.h>

int main (void)
{
    int i, j;
    bool locker[101];           // locker open = true, locker closed = false

    // student 1 opens all lockers

    for ( i = 1; i <= 100; ++i )
        locker[i] = true;

    // subsequent students toggle subsequent lockers

    for ( i = 2; i <= 100; ++i ) 
        for ( j = i; j <= 100; j += i )
            locker[j] = ! locker[j];

    // display results

    printf ("\nopen lockers: ");

    for ( i = 1; i <= 100; ++i )
        if ( locker[i] )
            printf ("%i ", i);

    printf ("\n");

    return 0;
}

打开的储物柜是完全平方数——任何具有奇数个因数的数字都是完全平方数。

Answer 5

不使用数组的解决方案。

#include <iostream>

using namespace std;

int main()
{

int studentTotal , lockerTotal,  visit, totalOpened = 0, totalClosed = 0;


cout << "Enter number of students" << endl;
cin >> studentTotal;



lockerTotal = studentTotal;

    for (int locker = 1; locker <= lockerTotal;  locker++ ){ // locker loop
        cout << "\n\n\nLocker no." << locker << endl;
        cout << " is visited by student(s) ";
        visit = 0;
        for (int student = 1 ; student <= studentTotal; student++) { // student loop

        if( locker % student == 0) {
                cout << student << ", ";
                visit++;}

        }//end of locker loop

        cout << "\nTotal number of visits: " << visit;
        if (visit % 2 == 0){
            cout << " the locker will stay closed.";
            totalClosed++;}
        else { cout << " the locker will be opened.";
            totalOpened++;}

    } //end of student loop

    if (lockerTotal == totalOpened + totalClosed) {
        cout << "\n\n\nOf total lockers (" << lockerTotal << "), " << totalOpened << " will be left open." << "(" << totalClosed << ") " << "will be closed." << endl;
        }else cout << "Error!!";



    return 0;

}

学生储物柜拼图

Student Locker puzzle

c

puzzle