日常训练2025-1-2

日常训练2025-1-2

D. Digital string maximization

rating:1300

https://codeforces.com/contest/2050/problem/D

思路：找trick + 暴力

根据题意，一个数值为 i 的数最多向左移动 i 步。而最大的数为 9 ，也就是说对于一个位置 x ，他的值只可能来自于 \([x, x + 9]\) 这个范围，更远的那些数不可能移动过来

所以即使是暴力时间复杂度也不过是\(O(9*n)\)​

评述

好烦这道题！！

代码

#include <bits/stdc++.h>

typedef std::pair<int, int> pii;
#define INF 0x3f3f3f3f
#define MOD 998244353
using i64 = long long;
const int N = 1e5+5;

void solve(){
	std::string s;
	std::cin >> s;

	int n = s.size();

	for (int i = 0; i < n; i++){
		int cur = s[i] - '0';
		int pos = -1;
		for (int j = i; j < n && j < i + 10; j++){
			if (s[j] - '0' - j + i > cur){
				pos = j;
				cur = s[j] - '0' - j + i;
			}
		}
		if (pos != -1){
			for (int k = pos; k >= i + 1; k--){
				s[k]--;
				std::swap(s[k], s[k-1]);
			}
		}
	}

	std::cout << s << '\n';
}

signed main()
{
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(2);
	int t = 1, i;
	std::cin >> t;
	for (i = 0; i < t; i++){
		solve();
	}
	return 0;
}

B. pspspsps

rating:1300

https://codeforces.com/problemset/problem/2049/B

思路（找trick）

直接找规律吧，没啥其他方法

代码

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;

#if !defined(ONLINE_JUDGE) && defined(LOCAL)
#include "helper.h"
bool IS_LOCAL = true;
#else
#define dbg(...) ;
bool IS_LOCAL = false;
#endif

void go() {
    int n;
    string s;
    cin >> n >> s;
    if (n == 1) {
        cout << "YES" << endl;
    } else if (!s.contains('s') || !s.contains('p')) {
        cout << "YES" << endl;
    } else if (s[0] == 's' && !s.substr(1, s.size() - 1).contains('s')) {
        cout << "YES" << endl;
    } else if (s[n - 1] == 'p' && !s.substr(0, n - 1).contains('p')) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    if (IS_LOCAL)
        std::cout << std::unitbuf;
    for (int batch = 1;; batch++) {
        int caseCnt;
        cin >> caseCnt;
        if (!caseCnt || cin.fail())
            break;
        for (int i = 1; i <= caseCnt; i++) {
            dbg(batch, i);
            go();
        }
    }
    return 0;
}