2.CodeForces 1935A Entertainment in MAC2024-06-16

3.CodeForces 1992C Gorilla and Permutation2024-07-16 4.CodeForces 1992D Test of Love2024-07-16 5.CodeForces 1992B Angry Monk2024-07-16 6.CodeForces 1992A Only Pluses2024-07-16 7.CodeForces 1983B Corner Twist2024-07-16 8.CodeForces 1983A Array Divisibility2024-07-16 9.CodeForces 1983C Have Your Cake and Eat It Too2024-07-16 10.CodeForces 1883B Chemistry2024-07-26 11.CodeForces 1883A Morning2024-07-26 12.CodeForces 1883C Raspberries2024-07-28 13.CodeForces 1883D In Love2024-07-28 14.CodeForces 1883E Look Back2024-07-28 15.CodeForces 1883F You Are So Beautiful2024-07-28 16.CodeForces 1883G1 Dances (Easy version)2024-07-28 17.CodeForces 908B New Year and Buggy Bot2024-08-01 18.CodeForces 908C New Year and Curling2024-08-01 19.CodeForces 1132B Discounts2024-08-01 20.CodeForces 1619D New Year's Problem2024-08-01 21.Codeforces Round 964 (Div. 4)2024-08-07

题目链接：Entertainment in MAC

思路

当当前操作次数n为偶数时，若原字符串大于反转字符串则可以将原字符串反转n - 2次，则得到的还是原字符串，此时反转一次，并将其再次反转的字符串加到反转字符串的末尾，此时得到词典最小的字符串，源字符串小于反转字符串时，直接将原字符串反转n次得到词典最小的字符串。
当当前操作次数为奇数时，若原字符串大于反转字符串，则将字符串反转n次，得到原串的反转字符串，即为词典最小的字符串，若原字符串小于反转字符串，则反转n - 1次，再将
当前字符串反转加到当前字符串末尾得到词典最小的字符串。

代码

 #include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 10;
 
 
void solve() {
    ll n;
    string s;
    cin >> n >> s;
    int l = 0, len = s.size();
    while (s[l] == s[len - 1 - l] && l < len / 2) {
        l++;
    }
    if (s[l] > s[len - l - 1]) {
        if (n % 2) {
            reverse(s.begin(), s.end());
        }
        else {
            string auxious = s;
            reverse(s.begin(), s.end());
            s = s + auxious;
        }
    }
    else if (s[l] < s[len - 1 - l]){
        if (n % 2) {
            string auxious = s;
            reverse(s.begin(), s.end());
            s = auxious + s;
        }
    }
    cout << s << endl;
}
int main(){
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

posted @ 2024-06-16 13:44 薛定谔的AC 阅读(7) 评论(0) 编辑收藏举报

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CodeForces 1935A Entertainment in MAC

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	#include<bits/stdc++.h>
	using namespace std;
	#define ll long long
	const int N = 1e5 + 10;


	void solve() {
	ll n;
	string s;
	cin >> n >> s;
	int l = 0, len = s.size();
	while (s[l] == s[len - 1 - l] && l < len / 2) {
	l++;
	}
	if (s[l] > s[len - l - 1]) {
	if (n % 2) {
	reverse(s.begin(), s.end());
	}
	else {
	string auxious = s;
	reverse(s.begin(), s.end());
	s = s + auxious;
	}
	}
	else if (s[l] < s[len - 1 - l]){
	if (n % 2) {
	string auxious = s;
	reverse(s.begin(), s.end());
	s = auxious + s;
	}
	}
	cout << s << endl;
	}
	int main(){
	int t;
	cin >> t;
	while (t--) {
	solve();
	}
	return 0;
	}