1 条题解

  • 0
    @ 2025-10-10 15:45:27

    C :

    #include<stdio.h>
    
    int fun(int n){
    	int r=1,i;
    	for(i=2;i*i<=n;i++){
    		if(n%i==0){
    			r=0;
    			break;
    		}
    	}
    	if(n==1){
    		r=0;
    	}
    	return r;
    }
    int tong(int n){
    	int r=0,i;
    	for(i=2;i*i<=n;i++){
    		if(n%i==0&&fun(i)==1&&fun(n/i)==1){
    			r=1;
    			break;
    		}
    	}
    	return r;
    }
    
    void main(){
    	int n;
    	scanf("%d",&n);
    
    	if(tong(n)==1){
    		printf("%s","It's a Tongtong number.");
    	}else{
    		printf("%s","It's not a Tongtong number.");
    		}
    }
    
    

    C++ :

    #include <iostream>
    #include <cmath>
    using namespace std;
    bool sushu(int n){
    	bool r=true;
    	int i;
    	for(i=2;i<=sqrt(n);i++){
    		if(n%i==0){
    			r=false;
    			break;
    		}
    	}
    	if(n<=1){
    		r=false;
    	}
    	return r;
    }
    void chaifeng(int n){
    	int i;
    	for(i=2;i<=n/2;i++){
    		if(n%i==0){
    			if(sushu(i)==true&&sushu(n/i)==true){ 
    			cout<<"It's a Tongtong number."<<endl;	
    			break;
    		}else {
    			cout<<"It's not a Tongtong number."<<endl;
    			break;
    			
    		}
    }
    	}
    }
    int main(){
    	int n;
    	cin>>n;
    	chaifeng(n);
    }
    

    Pascal :

    var
    n,i,x:longint;
    p:boolean;
    function ss(n:longint):boolean;
    var
    i:longint;
    begin
    if n<=1 then exit(false);
    for i:=2 to trunc(sqrt(n)) do
    if n mod i=0 then exit(false);
    exit(true);
    end;
    begin
    readln(n);
    for i:=2 to trunc(sqrt(n)) do
    if n mod i=0 then begin x:=n div i;if ss(x) then begin write('It','''','s a Tongtong number.');p:=true;break; end; end;
    if not(p) then write('It','''','s not a Tongtong number.');
    end.
    

    Java :

    import java.util.Scanner;
    
    public class Main{
    	public static void main(String[] args) {
    		int n=new Scanner(System.in).nextInt();
    		int count=0;
    		for(int i=1;i<=n;i++){
    			for(int j=1;j<=n;j++){
    				if(i*j==n&&isZs(i)&&isZs(j)){
    					count++;
    				}
    			}
    		}
    		if(count>0){
    			System.out.println("It's a Tongtong number.");
    		}else{
    			System.out.println("It's not a Tongtong number.");
    		}
    	}
    	public static boolean isZs(int n){
    		int count=0;
    		for(int i=1;i<=n;i++){
    			if(n%i==0&&i!=1&&i!=n){
    				count++;
    			}
    		}
    		if(count>0){
    			return false;
    		}else{
    			return true;
    		}	
    	}
    }
    

    Python :

    import math
    def suShu(n):
        f = True
        for i in range (2, int(math.sqrt(n))+1):
            if n % i == 0:
                f = False; 
        return f and n != 1
        
    n = int(input())
    f = False;
    for a in range(2,int(math.sqrt(n))+1):
            for b in range (int(math.sqrt(n))+1, n ):
                    if n == a * b and suShu(a) and suShu(b):
                        f = True;
    if f:
         print("It's a Tongtong number.")
    else:
         print("It's not a Tongtong number.")
    
    
    
    • 1

    信息

    ID
    146
    时间
    1000ms
    内存
    512MiB
    难度
    10
    标签
    递交数
    5
    已通过
    3
    上传者