1 条题解

  • 0
    @ 2025-10-10 15:45:27

    C :

    #include<stdio.h>
    #include<math.h>
    int find(int x){
    	int sum=0,i,t=sqrt(x);
    	for ( i=2; i<=t; i++)
    		if (x%i==0) sum+=i+x/i;
    	if (t*t==x) sum-=t;
    	return sum;
    }
    
    int main(){
    	int n,a,b,c;
    	scanf("%d",&n);
    	for ( a=2; a<=n; a++){
    		b=find(a); c=find(b);
    		if (a==c && a!=b && b<=n)
    		 printf("%d %d\n",a,b);
    	}
    	return 0;
    }
    

    C++ :

    #include <iostream>
    #include <cmath>
    using namespace std;
    int sum(int n){
    	int s,i;
    	s=0;
    	for(i=2;i<=sqrt(n);i++){
    		if(n%i==0){
    			if(i==n/i){
    				s=s+i;
    			}else{
    				s=s+i+n/i;
    			}
    		}
    	}
    	return s;
    }
    void shudui(int n){
    	int i,s;
    	for(i=2;i<=n;i++){
    		s=sum(i);
    		if(sum(s)==i&&i!=s&&s<=n){
    			cout<<i<<" "<<s<<endl;
    		}
    	}
    }
    int main(){
    	int n;
    	cin>>n;
    	shudui(n);
    }
    

    Pascal :

    var
        a:array[0..200000] of longint;
        i,j,n,s:longint;
    begin
        readln(n);
        for i:=2 to 2000 do
        begin
            s:=0;
            for j:=2 to i-1 do
                if i mod j=0 then inc(s,j);
            a[i]:=s;
        end;
        for i:=2 to n do
            if (a[a[i]]=i)and(a[i]<>i)and(a[i]<=n) then writeln(i,' ',a[i]);
    end.
    

    Java :

    import java.util.Scanner;
    public class Main {
    	public static void main(String[] args) {
    		Scanner sc = new Scanner(System.in);
    		int n = sc.nextInt();
    		qinMi(n);
    		
    	}
    	public static int forSum(int n) {
    		int sum = 0;
    		for(int i = 2; i <= n / 2; i++ ) {
    			
    			if(n % i == 0) {
    				sum += i;
    			}
    		}
    		return sum;
    		
    	}
    	
    	public static void qinMi(int n) {
    		int sum1 = 0;
    		int sum2 = 0;
    		for(int i = 2; i <= n ; i++ ) {
    			sum1 = forSum(i);
    			for(int k= 2; k <= n ; k++ ) {
    				sum2 = forSum(k);
    				if(sum1 == k && i == sum2 && i != k) {
    					System.out.println(i + " " + k);
    				}
    			}
    		}
    	}
    	
    	
    }
    

    Python :

    def yinZiHe(n):
        s=0
        for i in range(2,n):
            if n%i==0:
                s+=i
        return s
    n=int(input())
    for i in range(2,n+1):
           # i的因子和等于b b的因子和等于i
        if yinZiHe(yinZiHe(i))==i and yinZiHe(i)!=i and yinZiHe(i)<=n:
            print(i,yinZiHe(i))
    
    • 1

    信息

    ID
    135
    时间
    1000ms
    内存
    512MiB
    难度
    10
    标签
    递交数
    3
    已通过
    1
    上传者