• Codeforces Round 905 (Div. 3)
  • A. Morning
  • B. Chemistry
  • C. Raspberries

Codeforces Round 905 (Div. 3)

A. Morning

模拟光标移动即可

void solve(){
	string ss;
	cin >> ss;
	char ch = '1';
	int ans = 0;
	for(auto c : ss){
		if(c != ch){
			int x = c, y = ch;
			if(c == '0') x = '9' + 1;
			if(ch == '0') y = '9' + 1;
			ans += abs(x - y);
			ch = c;
		}
		++ ans;
	}
	cout << ans << '\n';
	return ;
}

B. Chemistry

先统计每种字符出现的次数，再统计字符出现次数为奇数的个数 \(cnt\)，如果 $cnt \le k + 1 $ 则可以满足要求，反之不行

void solve(){
	int n, k;
	string ss;
	cin >> n >> k >> ss;
	vector<int> cnt(26, 0);
	for(auto c : ss){
		++ cnt[c - 'a'];
	}
	int shu[2] = {0, 0};
	for(int i = 0; i < 26; ++ i)
		if(cnt[i]) ++ shu[cnt[i] % 2];
	if(k >= shu[1] - 1) cout << "YES\n";
	else cout << "NO\n";
	return ;
}

C. Raspberries

分类讨论
k = 2，3，5 时，只有给定的数变成其的倍数才行，记录改变所需最小值即
k = 4 需特判，k = 4 = 2 * 2，我们

void solve(){
	int n, ans = inf, k, a, shu[2] = {0, 0};
	cin >> n >> k;
	for(int i = 0; i < n; ++ i){
		cin >> a;
		ans = min(ans, (k - a % k) % k);
		++ shu[a % 2];
	}
	if(k == 4){
		if(shu[1] >= 2){
			int t = 2 - min(2, shu[0]);
			ans = min(ans, t);
		}else if(shu[1] == 1){
			if(shu[0] > 1) ans = 0;
			else if(shu[0] == 1) ans = min(ans, 1);
			else ans = min(ans, 3);
		}else{
			if(shu[0] >= 2) ans = 0;
		}
	}
	cout << ans << '\n';
	return ;
}