Codeforces Round #305 (Div. 2) A. Mike and Fax 暴力回文串-优快云博客

本文提供了一个关于如何判断一个字符串是否可以由给定数量的相同长度回文串组合而成的代码实现，适用于字符串长度较小的情况。通过简单的循环和比较操作，实现了判断逻辑，适合初学者理解和实践。

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A. Mike and Fax

Time Limit: 20 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/548/problem/A

Description

While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.

$\text{[math]}$

He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.

He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.

Input

The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).

The second line contains integer k (1 ≤ k ≤ 1000).

Output

Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.

Sample Input

saba
2

Sample Output

NO

HINT

题意

给你个字符串，告诉你这个字符串是由k个相同长度的回文串构成的，判断是否正确

题解：

数据范围很小，暴力判断就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 200001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

string s;
int k;
int main()
{
    //test;
    cin>>s>>k;
    if(s.size()%k!=0)
    {
        cout<<"NO"<<endl;
        return 0;
    }
    int len=s.size()/k;
    for(int i=0;i<k;i++)
    {
        for(int j=0;j<len;j++)
        {
            if(s[i*len+j]!=s[i*len+len-j-1])
            {
                cout<<"NO"<<endl;
                return 0;
            }
        }
    }
    cout<<"YES"<<endl;
}