1. Judgment of "completion" (40 points)

The sum of all factors of a positive integer is equal to itself. Such a number is called a perfect number. For example: 1 + 2 + 3 = 6, then 6 is the complete number. Output all completions up to 100.

Reference code:

#include<bits/stdc++.h> //Include all library functions 
using namespace std;

int isWholeNUM(int n){  //Use function to judge 
    int sum = 0;
    for(int i = 1;i < n;++i){
        if(n%i == 0)
            sum += i;
    }
    if(sum == n) return 1;
    else return 0;
}

int main(){
    for(int i = 1;i < 100;++i){
        if(isWholeNUM(i))
            cout<<i<<endl;
    }
    return 0;
} 

2. Sequence output (40 points)

Assume that the sequence satisfies:

1. The first digit is 1;
2. The second digit is 2;
3. The odd digit is the sum of the first two terms, and the even digit is the difference between the first two terms;

Sort and output the sequence within 100 in ascending order

Reference code:

#include<bits/stdc++.h>
using namespace std;

int r_list(int n){
    if(n == 1) 
        return 1;
    else if(n == 2)
        return 2;
    else if(n%2 != 0)
        return r_list(n - 1) + r_list(n -2);
    else if(n%2 == 0)
        return r_list(n - 1) - r_list(n -2);
}

int main(){
    int num;
    vector<int> array;
    for(int i = 1;i <= 100;++i){
        num = r_list(i);
        array.push_back(num);
    }
    sort(array.begin(),array.end());
    for(int i = 0;i < array.size();++i)
        cout<<array[i]<<endl;
    return 0;
} 

3. Shortest section (40 points)

There are 100 points on the plane. Any two points can be connected into a line segment. If there is a point (x1,y1) and a point (x2,y2), the distance between the two points is

Find the shortest of all line segments

Reference code:

#include<bits/stdc++.h>
using namespace std;

struct point{
    int x;
    int y;
} array[99];    //Contains 100 points, starting at 0 

int main(){
    int length = 0,max = 0;
    for(int i = 0;i < 100;++i)          //Enter coordinates for 100 points 
        cin>>array[i].x>>array[i].y;
    for(int i = 0;i < 100;++i){         //Calculate the length of all combined segments between 100 points 
        for(int j = 0;j < 100;++j){
            length = (array[i].x - array[j].x)^2 + (array[i].y - array[j].y)^2;
            if(max < length)
                max = length;
        }
    } 
    cout<<sqrt(max);
    return 0;
}

4. Number of 0 strings (30 points)

Input a sequence. Two or more consecutive zeros are called zero strings. If it is 001000, the number of 0 strings is 2. The input ends with 2 and asks if there are two or more 0 strings in the sequence

#include<bits/stdc++.h>
using namespace std;

int main(){
    string num;
    int m = 0,n = 0;
//  While ((CIN > > Num)! ='2 '); / / the question should be this way. When inputting 2, the input will be terminated automatically, but an error will be reported 
    cin>>num;                   //So manual input 
    for(int i = 0;i < num.length();++i){
        if(num[i] == '0')
            m++;
        else 
            m = 0;
        if(m >= 2 && num[i + 1] != '0')
            n++;    
    }
    if(n >= 2)
        cout<<"OK"<<endl;
    else
        cout<<"No"<<endl;
    return 0;
}