我就看谁敢抄

b = int(input())
a = [11,101,131,151,181,191,313,353,373,383,727,757,787,797,919,929,10301,10501,10601,11311,11411,12421,12721,12821,13331,13831,13931,14341,14741,15451,15551,16061,16361,16561,16661,17471,17971,18181,18481,19391,19891,19991,30103,30203,30403,30703,30803,31013,31513,32323,32423,33533,34543,34843,35053,35153,35353,35753,36263,36563,37273,37573,38083,38183,38783,39293,70207,70507,70607,71317,71917,72227,72727,73037,73237,73637,74047,74747,75557,76367,76667,77377,77477,77977,78487,78787,78887,79397,79697,79997,90709,91019,93139,93239,93739,94049,94349,94649,94849,94949,95959,96269,96469,96769,97379,97579,97879,98389,98689]
for i in a:
    if i < b:
        print(i)

C++ :

#include <iostream>
#include <cstdio>
#include <cmath>
#define Maxn 1009
using namespace std;
int dat[Maxn],len0=0,n;
bool isPrim(int m);
int main()
{
    scanf("%d",&n);
    dat[len0]=11;
    for(int i1=1; i1<=9; i1=i1+2){
        for(int i2=0; i2<=9; i2++){
            int m=i1*101+i2*10;
            if(isPrim(m)){
                dat[++len0]=m;
            }
        }
    }

    for(int i1=1; i1<=9; i1=i1+2){
        for(int i2=0; i2<=9; i2++){
            int m=i1*1001+i2*110;
            if(isPrim(m)){
                dat[++len0]=m;
            }
        }
    }
    for(int i=0; i<=len0; i++){
        if(dat[i]<=n){
            printf("%d\n",dat[i]);
        }else{
            break;
        }
    }
    return 0;
}

bool isPrim(int m)
{
    bool flag=true;
    int k=sqrt(m)+0.05;
    for(int i=2; i<=k; i++){
        if(m%i==0){
            flag=false;
            break;
        }
    }
    return flag;
}

本人不是很喜欢数学方法判断回文数，而是喜欢字符串的翻转（包括在C++中），尽管字符串的操作需要较高的时间复杂度

def is_prime(n):
    for i in range(2,int(n**0.5)+1):
        if n%i == 0:
            return False
    return True

def is_huiwen(n):
    s = str(n)
    if s == s[::-1]:
        return True
    else:
        return False

n = int(input())
for i in range(10,n+1):
    if is_huiwen(i) and  is_prime(i):
        print(i)

#include<bits/stdc++.h>
using namespace std;
bool isPalindrome(int x) {
    long revertedNumber = 0; 
    long t = x;
    if(x < 0 || (x % 10 == 0 && x != 0)) 
            return false;//排除小于0与末尾为0的情况
    while(t)
    {
        revertedNumber=10*revertedNumber+ (t % 10);
        t /= 10;
    }//将数倒转（把这个理解了基本上就做出来了）
        return x == revertedNumber;//比较是否相同
}//回文数判断
int main()
{
    int n;
    cin>>n;
	int a[100001]={0};
	for(int i=2;i<=n;i++)
	{
		a[i]=i;
	}
	for(int i=2;i<=n;i++)
	{
		for(int j=2;j<=sqrt(i);j++)
		{
			if(i%j==0)
			{
				a[i]=0;
				break;
			}
			
		}
		if(a[i]!=0)//质数判断
		{
			if(isPalindrome(a[i])&&a[i]>10)cout<<a[i]<<endl;
		}//回文数均大于10
	}
}

x=int(input())
def dj(x):
    r=True
    for a in range(2,int(x**0.5)+1):
        if x%a==0:
            r=False
            break
    return(r)
def op(i):
    s=str(i)[::-1]
    if s==str(i):
        return(True)
    else:
        return(False)
for i in range(10,x+1):
    if dj(i) and op(i):
        print(i)

def zxw(x):
    for i in range(2,int(x**0.5)+1):#缩时间
        if x%i==0:
            return 0
    s=str(x)
    if s==s[::-1]:#转字符串类型判断回文；再次感叹Python内置包多就是方便
        return 1
    return 0
n=int(input())
for j in range(11,n):
    if zxw(j)==True:
        print(j)

def check(x):
    if x==1 or x==0:
        return "NO"
    else:
        for i in range(2,int(x**0.5)+1):
            if x%i==0:
                return "NO"
        return "YES"
def huiwen(x):
    a=str(x)
    for i in range(len(a)//2):
        if a[::-1]==a:
            return 1
n=int(input())
for i in range(11,n+1):
    if check(i)=="YES"and huiwen(i)==1:
        print(i)

def ss(x):
    sum = True
    for m in range(2, x):
        if x % m == 0:
            sum = False
            break
    return sum
def hws(a):
    b = str(a)
    if b[::] == b[::-1]:
        return True
    else:
        return False
m = int(input())
for i in range(10,m+1):
    if ss(i) and hws(i):
        print(i)

python的哦~

n=int(input())
#n=200
x=[]
for i in range(2,n):
    t=1
    b=int(i**0.5)
    for k in range(2,b+1):
        if i%k==0:
            t=0
            break
    while t==1:
         x.append(i)
         break
for i in x:
    i=str(i)
    y=i[::-1]
    if i==y and len(i)>=2:
        print(i)

火影手游回调秽土鼬 n=0 b="" c="" m="" x=int(input()) for a in range(10,x+1): for i in range(1,a+1): if a%i0: n=n+1 m=str(a) if n2: for j in range(0,int(len(m)/2)+1): b=m[j]+b c=m[-j-1]+c if b==c: print(a) n=0 b="" c=""