C++ 解法, 执行用时: 4ms, 内存消耗: 392KB, 提交时间: 2022-01-09

#include <iostream>
#include <cmath>
using namespace std;
const int N=10010;
int q[N];
bool st[N];
void get_prime(int n)
{
	int j,i,cnt;
	cnt=0;
	for(i=2;i<=n;i++)
	{
		if(!st[i])
		{
			q[cnt++]=i;
			for(j=i+i;j<=n;j+=i)
				st[j]=true;
		}
	}
}
bool isprime(int x)
{
	int i;
	int sa=sqrt(x);
	for(i=2;i<=sa;i++)
	{
		if(x%i==0) return false;
	}
	return true;
}
int main(){
	int a,i,b;
	cin>>a>>b;
	if(a>b)
	{
		int t;
		t=b;
		b=a;
		a=t;
	}
	//get_prime(N);
	int cnt=0;
	st[1]=true;
	for(i=a;i<=b;i++)
	{ 
		if(i<=2);
		else if(isprime(i)) cnt++;
	}
	printf("%d到%d之间有%d个大于2的素数",a,b,cnt);
	return 0;
}

C++ 解法, 执行用时: 25ms, 内存消耗: 428KB, 提交时间: 2022-01-09

#include <stdio.h>
int fact(int);
int main(void){
    int m, n, max  = 0, min = 0, i = 0,l = 0,s = 0;
    scanf("%d %d", &m, &n);
    max = m > n ? m : n;
    min = m < n ? m : n;
    for(i = min; i <= max; i ++){
        l = fact(i);
        if(l) s ++;
    }
    printf("%d到%d之间有%d个大于2的素数",min,max,s);
    return 0;
}
int fact(int n){
    int i = 2;
    for(; i < n; i ++){
        if(n % i == 0) break;
    }
    if(i == n) return 1;
    else return 0;
}

Java 解法, 执行用时: 29ms, 内存消耗: 10700KB, 提交时间: 2021-11-17

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int start = scanner.nextInt();
        int end = scanner.nextInt();
        method(start,end);
    }
    
    public static boolean isPeime(int num){
        for(int i = 2; i * i <= num; i++){
            if(num % i == 0) return false;
        }
        return true;
    }

    public static void method(int start, int end) {
        int count=0;

        //write your code here......
        if(start > end){
            int change = start;
            start = end;
            end = change;
        }
        
        for(int i = start; i  <= end; i++){
            if(i <= 2) continue;
            if(isPeime(i)) count ++;
        }

        System.out.println(start+"到"+end+"之间有"+count+"个大于2的素数");
    }
}

Java 解法, 执行用时: 30ms, 内存消耗: 10520KB, 提交时间: 2022-02-10

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int start = scanner.nextInt();
        int end = scanner.nextInt();
        method(start,end);
    }

    public static void method(int start, int end) {
        int count=0;

        //write your code here......
        if(start>end){
            int temp;
            temp=start;
            start=end;
            end=temp;
        }
        for(int i=start;i<=end;i++){
            if(i>2&&issushu(i)){
                count++;
            }
        }        
        System.out.println(start+"到"+end+"之间有"+count+"个大于2的素数");
    }
      public static boolean issushu(int num){        
        for(int i=2;i*i<=num;i++){
            //只要能被i整除，则不是素数
            if(num%i==0){
                return false;
            }
        }
        return true;
        }
}

Java 解法, 执行用时: 30ms, 内存消耗: 10608KB, 提交时间: 2022-02-09

import java.util.Scanner;

public class Main {
     public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int start = scanner.nextInt();
        int end = scanner.nextInt();
        method(start, end);
    }

    public static boolean isPeime(int num) {
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                return false;
            }

        }
        return true;
    }

    public static void method(int start, int end) {
        int count = 0;
        if (start > end) {
            int temp = start;
            start = end;
            end = temp;
        }
        for (int i = start; i <= end; i++) {
            if (i <= 2) {
                continue;
            }
            if (isPeime(i)) {
                count++;
            }
        }
            System.out.println(start + "到" + end + "之间有" + count + "个大于2的素数");


      
    }
}