Educational Codeforces Round 29 - A. Quasi-palindrome

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本文介绍了一个算法问题，即如何判断一个整数是否为准回文数。通过去除数字尾部的0并检查剩余部分是否为回文来解决这个问题。

Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.

String t is called a palindrome, if it reads the same from left to right and from right to left.

For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.

You are given some integer number x. Check if it's a quasi-palindromic number.

Input

The first line contains one integer number x (1 ≤ x ≤ 109). This number is given without any leading zeroes.

Output

Print "YES" if number x is quasi-palindromic. Otherwise, print "NO" (without quotes).

Examples

input
131

output
YES

input
320

output
NO

input
2010200

output
YES

题意：

给你一个数字，你可以在数字最前或最后添加任意个数的0。问你操作之后能不能使数字变成回文。

POINT：

把尾0去掉，接着判断回文就行

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include<algorithm>
using namespace std;

int main()
{
    int n;
    scanf("%d",&n);
    while(n%10==0) n=n/10;
    int s[22];
    int cnt=0;
    while(n){
        s[++cnt]=n%10;
        n/=10;
    }
    int ans=1;
    for(int i=1;i<=cnt/2;i++){
        if(s[i]!=s[cnt-i+1]){
            ans=0;
            break;
        }
    }
    if(ans) printf("YES\n");
    else printf("NO\n");
}