Huawei trial (7)

Intermediate question: 100 points
Description: an N (1 ≤ n ≤ 65535) electric light is installed in the corridor, numbers 1, 2, 3 ,... N-1 and N. Each lamp is controlled by a wire pulling switch. Start, all the lights are turned off.
N students passed through the corridor. The first student pulls the switch of the electric lamp whose number is a multiple of 1, and the second student pulls the switch of the electric lamp whose number is a multiple of 2; then the third student pulls the switch of the electric lamp whose number is a multiple of 3. Then, the nth student pulls the switch of the electric lamp whose number is a multiple of N. After N students finished according to this rule, there were several lights in the corridor. Note: The number of electric lights is the same as that of students.

 
Input: number of electric lights
Output: Number of lights on
Sample input: 3, sample output: 1

Solution 1: If you want to simulate this question, it is certainly possible, but it is hard for students with poor code Foundation to write it. Here, we will write a simulation method.
Solution 2: think more about this question. We can see that for any lamp, such as 12, the factors include 1, 2, 3, 4, 6, and 12. Note: Students numbered 1, 2, 3, 4, 6, and 12 need to pull it. Here we find that 1*12 = 12, 2*6 = 12, 3*4 = 12, so the factors are all one-to-one, that is, the 1 pull is offset by 12, and the 2 pull is offset by 6. So who will not be offset? The answer is the number of percentages. For example, if the factor of 9 is or 9, then 3*3 = 9, and 3 is pulled only once, it will not be offset. Therefore, the answer to this question is that the number of records within the input N is equal to (INT) SQRT (n ).
From this question, we can see that in many programming cases, the thinking test is much more useful than the hyperactivity pen.

#include<iostream>using namespace std;void main(){    int n;    cin>>n;    int numLightOn = 0;    for(int i=1;(i*i)<=n;i++)       numLightOn++;    cout<<numLightOn<<endl;  }

The simulated code is as follows:

# Include <iostream> # include <vector> using namespace STD; void main () {int N; CIN> N; vector <int> light (n + 1, 0 ); // 0 indicates extinct, and 1 indicates bright int numlighton = 0; For (INT I = 1; I <= N; I ++) // n people {Int J = I; int times = 2; while (j <= N) {Light [J] = 1-light [J]; // 0-> 1, 1-> 0 J = I; // The J lamp J * = times; times ++ ;}} // end for (INT I = 1; I <= N; I ++) {If (light [I] = 1) numlighton ++ ;} cout <numlighton <Endl ;}