Educational Codeforces Round 168 (Rated for Div. 2)

合集 - 记录随手刷过的编程题(76)

1.E. Arranging The Sheep2024-02-282.C. Team2024-02-293.C. RationalLee2024-03-014.B. Preparing Olympiad2024-03-055.B. Find The Array2024-03-056.E. Accidental Victory2024-03-067.A. Learning Languages2024-03-078.A. k-th divisor2024-03-089.B. Quasi Binary2024-03-0910.蓝桥杯-地宫取宝2024-03-0911.D. Divide by three, multiply by two2024-03-1212.D. Game With Array2024-03-1313.B. Composite Coloring2024-03-1414.洛谷 P1204 [USACO1.2] 挤牛奶Milking Cows2024-04-2015.洛谷 P3353 在你窗外闪耀的星星2024-04-2216.洛谷 P8818 [CSP-S 2022] 策略游戏2024-04-2317.Educational Codeforces Round 164 (Rated for Div. 2)2024-04-2518.E. Chain Reaction2024-04-2619.Manthan, Codefest 18 (rated, Div. 1 + Div. 2) D. Valid BFS?2024-04-2920.D. Distance in Tree2024-04-2921.D. Multiplication Table2024-04-3022.C. Mixing Water2024-05-0123.B. Greg and Graph2024-05-0624.D. Another Problem About Dividing Numbers2024-05-0725.Acwing 143. 最大异或对2024-05-0726.Acwing 3485. 最大异或和2024-05-0727.C. Checkposts2024-05-0828.E. Lomsat gelral2024-05-1129.D. Powerful array2024-05-1530.D. Deleting Divisors2024-05-1631.Codeforces Round 950 (Div. 3)2024-06-0432.Codeforces Round 951 (Div. 2)2024-06-0733.SuntoryProgrammingContest2024（AtCoder Beginner Contest 357）2024-06-0934.Codeforces Round 952 (Div. 4)2024-06-1235.AtCoder Beginner Contest 3582024-06-1736.B. Modulo Sum2024-06-2137.D. Soldier and Number Game2024-06-2238.AtCoder Beginner Contest 3592024-06-2439.Codeforces Round 954 (Div. 3)2024-06-2440.Codeforces Round 955 (Div. 2, with prizes from NEAR!)2024-06-2641.AtCoder Regular Contest 1802024-06-3042.D - Ghost Ants2024-07-0343.AtCoder Beginner Contest 361）2024-07-0744.Codeforces Round 957 (Div. 3)2024-07-1245.E. Decode2024-07-29

46.Educational Codeforces Round 168 (Rated for Div. 2)2024-08-06

47.B. Parity and Sum2024-08-0848.P1972 [SDOI2009] HH的项链2024-08-1249.Codeforces Round 966 (Div. 3) VP2024-08-1550.https://codeforces.com/contest/2004/problem/B2024-08-1951.C. Splitting Items2024-08-1952.D. Colored Portals2024-08-2053.Codeforces Round 967 (Div. 2)2024-08-2154.D. Determine Winning Islands in Race2024-08-2355.C. Turtle and Good Pairs2024-08-2856.Codeforces Round 970 (Div. 3)（VP）2024-09-0357.G. Sakurako's Task2024-09-0458.Codeforces Round 971 (Div. 4)2024-09-0459.C. Lazy Narek2024-09-1960.B. Regular Bracket Sequence02-2161.C. News Distribution02-2162.B. Karen and Coffee02-2163.G. 2^Sort02-2564.C. Johnny and Another Rating Drop03-0465.A. Greg and Array03-0566.D2. Magic Powder - 203-0667.C. Schedule Management03-0668.C. Constanze's Machine03-0769.C. Soldier and Cards03-0870.C. Berland Regional03-1071.E. Anna and the Valentine's Day Gift03-1172.B. Polo the Penguin and Matrix03-1273.B. Fortune Telling03-1374.C. Powers Of Two03-1375.D. Palindromes Coloring03-1476.C. Phoenix and Towers03-15

收起

没有时间参赛 直接补几道简单题吧~

B. Make Three Regions

题意：给一个2行的字符串，有block和其他东西，问把一个位置变成block让联通的部分变成3个部分，有多少种方法

思路：找规律，找所哟符合条件的块即可

void solve(){
	int n;
	cin >> n;

	array<string, 2> s;
	cin >> s[0] >> s[1];

	int ans = 0;
	for (int i = 1; i < n - 1; ++i) {
		for (int j = 0; j < 2; ++j) {
			if (s[j][i] == '.' && s[j][i - 1] == 'x' && s[j][i + 1] == 'x' &&
				s[j ^ 1][i] == '.' && s[j ^ 1][i - 1] == '.' && s[j ^ 1][i + 1] == '.'
			)  {
				ans ++;
			}
		}
	}

	cout << ans << '\n';
}

C. Even Positions

题意：给定长度为n的括号字符串，下标1~n，并且所有奇数位置上的字符让自己填。字符串的代价是匹配括号的距离。求所有填充字符的方法中最小的距离。

思路：一眼贪心，首先字符串要合法。在合法的前提下，肯定是如果有左括号就直接填右。如果没有左，就填左。

总结：这个算法的正确性证明比较有趣。 会不会出现按照这个策略填充，会出现到了最后左右括号数量不相等的情况？

void solve(){
	int n;
	cin >> n;
	string s;
	cin >> s;

	int cnt = 0;
	for (int i = 0; i < n; ++i) {
		if (s[i] == '_') {
			if (!cnt) {
				cnt ++;
				s[i] = '(';
			}
			else {
				s[i] = ')';
				cnt --;
			}
		}
		else {
			cnt += (s[i] == '(' ? 1 : -1);
		}
	}

	int ans = 0;
	for (int i = 0; i < n; ++i) {
		if (s[i] == '(') {
			ans -= i;
		}
		else {
			ans += i;
		}
	}

	cout << ans << '\n';
}

D. Maximize the Root

题意：给定一颗树，每个结点有权值。每次操作可以将以当前结点v的子树中，所有的子节点权值-1，然后v的权值+1。不限操作次数的操作后，root根的编号为1的最大权值是多少？

思路：操作过程中要求结点权值不能为负数，所以结点1的权值增加的幅度，就是树中所有的结点，经过操作后值最小的最大值。于是问题转化为，经过若干次操作后，权值最小的结点，最大是多少。 直接dfs遍历，统计当前子树中权值最小的结点，分两种情况讨论。如果当前结点权值<最小权值结点，则当前结点权值变大，否则不变。

总结：一开始理解错题了，以为权值减小的结点只跟与当前结点直接相连的子节点有关。后来tc过不去才发现是子树中所有的结点都要操作，丢。

void solve(){
    int n;
    cin >> n;
    
    vector<vector<int>> al(n + 1);
    vector<long long> weights(n + 1);
    for (int i = 1; i <= n; ++i) {
        cin >> weights[i];
    }
    
    for (int i = 2; i <= n; ++i) {
        int p;
        cin >> p;
        al[p].push_back(i);
    }

    auto dfs = [&](auto&& self, int u)->long long {
        long long x = INF_LL;
        for (const auto& v : al[u]) {
            x = min(self(self, v), x);
        }
        if (x!= INF_LL) {
            if (u == 1) {
                weights[u] += x;
            }
            else if (x > weights[u]) {
                weights[u] = (weights[u] + x) / 2;
            }
            else {
                weights[u] = x;
            }
        }
        return weights[u];
    };

    cout << (dfs(dfs, 1)) << '\n';
}

有模板写程序就是好用

https://github.com/yxc-s/programming-template