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  • 【codeforces 548A】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
    input
    saba
    2
    output
    NO
    input
    saddastavvat
    2
    output
    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”.

    【题目链接】:http://codeforces.com/contest/548/problem/A

    【题解】

    已经提示你每个串的长度都是一样的了;
    然后这个总串是由若干个这样的串排列组成(意思是说没有其他的字符);
    则每n/k个字符都应该是回文;
    判断一下就好;

    【完整代码】

    #include <bits/stdc++.h>
    using namespace std;
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define LL long long
    #define rep1(i,a,b) for (int i = a;i <= b;i++)
    #define rep2(i,a,b) for (int i = a;i >= b;i--)
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define rei(x) scanf("%d",&x)
    #define rel(x) scanf("%I64d",&x)
    
    typedef pair<int,int> pii;
    typedef pair<LL,LL> pll;
    
    //const int MAXN = x;
    const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
    const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
    const double pi = acos(-1.0);
    
    const int MAXN = 1e3+100;
    char s[MAXN];
    int len,k,le;
    
    bool is(int l,int r)
    {
        if (l>=r) return true;
        if (s[l]==s[r])
            return is(l+1,r-1);
        else
            return false;
    }
    
    int main()
    {
        //freopen("F:\rush.txt","r",stdin);
        scanf("%s",s+1);
        rei(k);
        len = strlen(s+1);
        if ((len%k)!=0)
        {
            puts("NO");
            return 0;
        }
        le = len/k;
        int l = 1,r = le;
        while (r<=len)
        {
            if (!is(l,r))
            {
                puts("NO");
                return 0;
            }
            r+=le,l+=le;
        }
        puts("YES");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7626786.html
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