CF1072A Palindromic Twist 思维

Palindromic Twist

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string $\mathrm{ss consisting\; of nn lowercase\; Latin\; letters. nn is\; even.}$

For each position $\mathrm{ii (1\le i\le n1\le i\le n)\; in\; string ss you\; are\; required\; to\; change\; the\; letter\; on\; this\; position\; either\; to\; the\; previous\; letter\; in\; alphabetic\; order\; or\; to\; the\; next\; one\; (letters\; \text{'}a\text{'}\; and\; \text{'}z\text{'}\; have\; only\; one\; of\; these\; options).\; Letter\; in\; every\; position\; must\; be\; changed exactly\; once.}$

For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.

That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' $\to \to \text{'}d\text{'},\; \text{'}o\text{'} \to \to \text{'}p\text{'},\; \text{'}d\text{'} \to \to \text{'}e\text{'},\; \text{'}e\text{'} \to \to \text{'}d\text{'},\; \text{'}f\text{'} \to \to \text{'}e\text{'},\; \text{'}o\text{'} \to \to \text{'}p\text{'},\; \text{'}r\text{'} \to \to \text{'}q\text{'},\; \text{'}c\text{'} \to \to \text{'}b\text{'},\; \text{'}e\text{'} \to \to \text{'}f\text{'},\; \text{'}s\text{'} \to \to \text{'}t\text{'}).$

String $\mathrm{ss is\; called\; a\; palindrome\; if\; it\; reads\; the\; same\; from\; left\; to\; right\; and\; from\; right\; to\; left.\; For\; example,\; strings\; "abba"\; and\; "zz"\; are\; palindromes\; and\; strings\; "abca"\; and\; "zy"\; are\; not.}$

Your goal is to check if it's possible to make string $\mathrm{ss a\; palindrome\; by\; applying\; the\; aforementioned\; changes\; to\; every\; position.\; Print\; "YES"\; if\; string ss can\; be\; transformed\; to\; a\; palindrome\; and\; "NO"\; otherwise.}$

Each testcase contains several strings, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer $\mathrm{TT (1\le T\le 501\le T\le 50)\; —\; the\; number\; of\; strings\; in\; a\; testcase.}$

Then $\mathrm{2T2T lines\; follow\; —\; lines (2i-1)(2i-1) and 2i2i of\; them\; describe\; the ii-th\; string.\; The\; first\; line\; of\; the\; pair\; contains\; a\; single\; integer nn (2\le n\le 1002\le n\le 100, nn is\; even)\; —\; the\; length\; of\; the\; corresponding\; string.\; The\; second\; line\; of\; the\; pair\; contains\; a\; string ss,\; consisting\; of nnlowercase\; Latin\; letters.}$

Output

Print $\mathrm{TT lines.\; The ii-th\; line\; should\; contain\; the\; answer\; to\; the ii-th\; string\; of\; the\; input.\; Print\; "YES"\; if\; it\text{'}s\; possible\; to\; make\; the ii-th\; string\; a\; palindrome\; by\; applying\; the\; aforementioned\; changes\; to\; every\; position.\; Print\; "NO"\; otherwise.}$

Example

input

Copy

5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml

output

Copy

YES
NO
YES
NO
NO

Note

The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.

The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.

The third string can be changed to "beeb" which is a palindrome.

The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".

 

 题意：每个字母可以向后向前一位或者保持不变，要变的时候必须同时变两个，问经过变化后是否可以得到回文字符串

分析：判断前后字符变化或者不变化后是否相等

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <windows.h>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
string s;
ll n;
bool ok(ll x) {
    //debug(s[x]), debug(s[n-x-1]);
    if( s[x] == s[n-x-1] || s[x]+1 == s[n-x-1]-1 || s[x]+1 == s[n-x-1]+1 || s[x]-1 == s[n-x-1]-1 || s[x]-1 == s[n-x-1]+1 ) {
        return true;
    }
    return false;
}
int main() {
    ios::sync_with_stdio(0);
    ll T;
    cin >> T;
    while( T -- ) {
        cin >> n;
        cin >> s;
        bool flag = true;
        ll t;
        if( n%2 != 0 ) {
            t = (n+1)/2-1;
        } else {
            t = (n+1)/2;
        }
        for( ll i = 0; i < t; i ++ ) {
            if( !ok(i) ) {
                flag = false;
                break;
            }
        }
        if( flag ) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }
    return 0;
}

　　

转载于:https://www.cnblogs.com/l609929321/p/9502756.html