CodeForces 1883B Chemistry

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-16 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：CodeForces 1883B【Chemistry】

思路

判断最多删去k个字符后剩下的部分为回文字符串，所以优先删除将个数为奇数个的相同字符删为偶数，当最后留下的字符串中，奇数个数的相同字符种类小于等于1时才会是回文字符串，如：aaabbbccc，此时个数为奇数的相同字符种类有三种，分别为a，b，c，所以最少要将a，b，c三种中的两种字符分别删去一个，才能得到回文字符串，假设删去一个b和一个c，得到aaabbcc，排列后得到cbaaabc。所以先计算奇数个数的相同字符种类小于等于k + 1时，就可以通过操作得到回文串。

代码

 #include <bits/stdc++.h>
using namespace std;
 
int num[30];
void solve() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
 
  memset(num, 0, sizeof num);
 
  int len = s.size();
  for (int i = 0; i < len; i++) {
    num[s[i] - 'a']++;
  }
 
  int count = 0;
  for (int i = 0; i < 26; i++) {
    if (num[i] % 2)
      count++;
  }
 
  if (count - k <= 1) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
}
 
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

posted @ 2024-07-26 21:23 薛定谔的AC 阅读(8) 评论(0) 编辑收藏举报

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CodeForces 1883B Chemistry

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

	int num[30];
	void solve() {
	int n, k;
	cin >> n >> k;
	string s;
	cin >> s;

	memset(num, 0, sizeof num);

	int len = s.size();
	for (int i = 0; i < len; i++) {
	num[s[i] - 'a']++;
	}

	int count = 0;
	for (int i = 0; i < 26; i++) {
	if (num[i] % 2)
	count++;
	}

	if (count - k <= 1) {
	cout << "YES" << endl;
	} else {
	cout << "NO" << endl;
	}
	}

	int main() {
	int t;
	cin >> t;
	while (t--) {
	solve();
	}
	}