A - Mike and Fax CodeForces - 548A（回文串）

最新推荐文章于 2021-05-29 15:44:01 发布

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最新推荐文章于 2021-05-29 15:44:01 发布

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原文链接：http://www.cnblogs.com/tomjobs/p/10612581.html

本文介绍了一个简单的算法，用于检查字符串是否由相同长度的k个回文子串组成。该算法通过逐个验证每个子串是否为回文来完成任务。

A. Mike and Fax
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

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.

Examples
inputCopy
saba
2
outputCopy
NO
inputCopy
saddastavvat
2
outputCopy
YES
Note
Palindrome is a string reading the same forward and backward.

In the second sample, the faxes in his back-bag can be “saddas” and “tavvat”.

思路：模拟判断回文串题目

//找回文串题目，难道要用EKMP或者什么鬼manacher??

#include <bits/stdc++.h>

using namespace std;

int check(int s,int e,string &str)
{
    int flag = 0;
    for(int i = s;i <= e;i++)
    {
        if(str[i] == str[e - (i - s)])
        {
            continue;
        }
        else
        {
            flag = 1;
            break;
        }
    }
    return flag;//1为不回文，0为回文。
}

int main()
{
    string str;
    while(cin>>str)
    {
        int k;
        cin>>k;
        if(str.length() % k)//这个判断暂时搁置
        {
            printf("NO\n");
            continue;
        }
        int t = str.length() / k;
        int flag = 1;
        for(int i = 0;i < str.length();i += t)
        {
            if(check(i, i + t - 1,str))
            {
                flag = 0;
                break;
            }
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
}