#include<stdio.h>
int main(){
 	int i,tmp,input,sum=0;
	scanf("%d",&input);
	for(i=input;i;i/=10){
		tmp=i%10;
		sum=sum*10+tmp;		//计算出倒数 
	}
    sum==input?printf("yes"):printf("no");
    return 0;
}

#include<stdio.h>
#include <string.h>
int main()
{
int lang;
char str[1000],str1[1000];
gets(str);
lang=strlen(str);
for (int i = 0; i < lang; i++)
{
str1[i] = str[lang - 1 - i];
}
//str[lang] = '\0';
str1[lang] = '\0';
/*printf("%s\n", str);
printf("%s", str1);
printf("%d", strcmp(str, str1));*/
if (strcmp(str, str1)==0)
{
printf("yes");
}
else
{
printf("no");
}
return 0;
}

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String n = sc.next();
        int first = 0, end = n.length() - 1;
        boolean isPalindrome = true;
        
        while (first < end) {
            if (n.charAt(first) == n.charAt(end)) {
                first++;
                end--;
            } else {
                isPalindrome = false;
                break;
            }
        }
        
        if (isPalindrome) {
            System.out.println("yes");
        } else {
            System.out.println("no");
        }
    }
}

双指针的算法思想，可以去了解了解，我这里是java写的

#include <stdio.h>
#include <stdlib.h>
#include "math.h"
int main()
{
    int nb(int a);
    int power(int a, int b);
    int a,n,i;
    scanf("%d",&a);
    n=nb(a);
    i=n;
    int num[n],b=a;
    while (i>0)
    {
        num[i]=(int)(b/(int)pow(10,i-1));
        b=b-num[i]*(int)pow(10,i-1);
        i--;
    }

    i=1;
    int c;
    while (i<=n)
    {
        c+=num[i]*pow(10,n-i+1)/10;
        i++;
    }
    if(c==a)printf("yes");
    else printf("no");

    system("pause");
    return 0;
}

int nb(int a)
{
    int nb=0,ns=1;
    while(a/ns!=0)
    {
        nb++;
        ns=10*ns;
    }
    return nb;
}//

懒得算.png

#include<iostream>
using namespace std;
int main() {
    string s;
    cin >> s;
    bool flag = 1;
    int longofNum = s.size();
    for (int i = 0;i < longofNum - 1;i++) {
        if (s[i] != s[longofNum - 1 - i]) flag = 0;
    }
    if (flag) cout << "yes";
    else cout << "no";
}

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{ 
 	int hws,b,sum=0;
 	cin>>hws;
 	for(int a=hws;a>0;a/=10)
 	{
 		b=a%10;//一次是个，2次十位，等等
		sum=sum*10+b; 
	 }
	 if(sum==hws)
	 {
	 	cout<<"yes";
	 
	 }
	 	else
	 	cout<<"no";
		return 0;
}
yujie好厉害

#include <stdio.h> int main() { int x,a,b=0,c; scanf("%d",&x); a=x; while(x!=0) { c=x%10; b=b*10+c; x/=10; } if(a==b) { printf("yes\n"); } else { printf("no\n"); } return 0; }

a=input()
if a[::-1]==a:
    print("yes")
else:
    print("no")

#include<stdio.h> int main() { int i,n,j=0; int a[100]; int k=0; scanf("%d",&n); while(n>0) { i=n%10; a[j]=i; n=n/10; j=j+1; } for(i=0;i<j/2;i++) { if(a[i]!=a[j-1-i]) { k++; } } if(k==0) { printf("yes"); } else { printf("no"); } return 0; }

#include<stdio.h> int isPalindrome(int x){ if(x<0){ return 0; } int div=1; while(x / div>=10){ div*=10; } while(x>0){ int left = x / div; int right = x % 10; if(left != right){ return 0; } x = (x % div)/10; div /= 100; } return 1; }

int main() { int x; scanf("%d",&x); if(isPalindrome(x)){ printf("yes"); }else{ printf("no"); }

return 0;

}

#include<stdio.h> int main(){ int X; int Y=0; scanf("%d",&X); for(int i=X;i>0;i=i/10){ Y=i%10+Y*10; } if(Y==X) printf("yes");

else
    printf("no");

return 0;

}

#include<iostream> using namespace std;

int main() { int n,b; cin>>n; int a=n; while(n>0) { b=10*b+n%10; n/=10; } if(a==b) { cout<<"yes"<<endl; } else { cout<<"no"<<endl; } return 0; }

#include<stdio.h> int main(){ int n,a=0; scanf("%d",&n); for(int i=n;i>0;i/=10) { a=a*10+i%10;} if(n==a){printf("yes");} else {printf("no");} return 0; }

#include<stdio.h> bool hws(long int x) {

long int s, y=0;
s=x;
while(s>0)
{
	y=y*10+s%10;
	s=s/10;
	
}
if(y==x)
return true;
else
return false;

} int main(){ long int n; scanf("%ld",&n); bool flag = hws(n); if(flag==0){ printf("no");}else{ printf("yes");} return 0;

}

#include<stdio.h> int main() { int j,n=0,X; scanf("%d",&X); j=X; while(j) { n=n*10+j%10; j/=10; } if(n==X) printf("yes"); else printf("no");

}