Codeforces Round 962 (Div. 3)

合集 - 题解(19)

1.[博客迁移20190713]题解 P4169 【[Violet]天使玩偶/SJY摆棋子】2024-05-272.模拟题2024-06-263.SP666 VOCV - Con-Junctions 题解2024-08-254.Codeforces Round #900 (Div. 3)2024-08-25

5.Codeforces Round 962 (Div. 3)2024-08-27

6.题解 CF993E 【Nikita and Order Statistics】2024-09-167.题解 P6891 【[JOISC2020]ビルの飾り付け 4】2024-09-188.P2163 [SHOI2007] 园丁的烦恼2024-09-189.[POI2011] LIZ-Lollipop2024-09-1910.Codeforces Round 973 (Div. 2)2024-09-2111.Codeforces Round 974 (Div. 3)2024-09-2212.NEERC2013题解2024-09-2313.CF1446C Xor Tree2024-09-2514.Codeforces Round 709 (Div. 1)2024-09-2715.Codeforces Round 975 (Div. 2)2024-09-2816.NEERC2014题解2024-09-2917.题解 P2726 【[SHOI2005]树的双中心】2024-10-0118.CF2019 F. Max Plus Min Plus Size2024-10-0319.Codeforces Round 757 (Div. 2)03-05

收起

A. Legs

若只判断题，根据模4意义下分类即可。

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B. Scale

模拟题，缩小同值矩阵即可。

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C. Sort

对每个字母求前缀数量和，答案就是两端点的差。

const int N = 2e5 + 5;
int T, n, q; char a[N], b[N]; int c[N][26], d[N][26];

signed main(void) {
	for (read(T); T; T--) {
		read(n), read(q);
		readstr(a + 1); readstr(b + 1);
		for (int i = 1; i <= n; i++) {
			for (int j = 0; j < 26; j++) c[i][j] = c[i - 1][j], d[i][j] = d[i - 1][j];
			c[i][a[i] - 'a']++; d[i][b[i] - 'a']++;
		}
		
		for (int l, r; q; q--) {
			read(l), read(r); int ans = 0;
			for (int i = 0; i < 26; i++) {
				int cnt = (c[r][i] - c[l - 1][i]) - (d[r][i] - d[l - 1][i]);
				if (cnt > 0) ans += cnt;
			} writeln(ans);
		}
	}
//	fwrite(pf, 1, o1 - pf, stdout);
	return 0;
}

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D. Fun

由两个不等式，在枚举 $a$ 的时候可以限制 $b$ 右端点，在确定两者的情况下可以直接算出 $c=min((n-a\ast b)/(a+b),x-(a+b))$

int T, n, x; ll ans;

signed main(void) {
	for (read(T); T; T--) {
		read(n), read(x); ans = 0;
		int m = min(n, x);
		for (int a = 1; a <= m; a++) {
			for (int b = 1; b * a < n && a + b < x; b++) {
				int c = min((n - a * b) / (a + b), x - (a + b));
				ans += c;
			}
		} writeln(ans);
	}
//	fwrite(pf, 1, o1 - pf, stdout);
	return 0;
}

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E. Decode

如何有效地检查一个数组是否包含相同数量的 0 和 1？其实就是其中一个的数量是区间长度的一半，将这个式子移项并平移，可以得到类似计数端点相同值的等价表达。

const int N = 2e5 + 5;
const int mod = 1e9 + 7;
int T, n, b[N], ans, c[2 * N]; char a[N];

//inline void update(int pos, int val) {
//	for ( ; pos <= 2 * n; pos += lowbit(pos)) c[pos] = (c[pos] + val) % mod;
//}
//
//inline int query(int pos) {
//	int ret = 0;
//	for (; pos; pos -= lowbit(pos)) ret = (ret + c[pos]) % mod;
//	return ret;
//}

signed main(void) {
	for (read(T); T; T--) {
		readstr(a + 1); n = strlen(a + 1); ans = 0;
		c[n + 1] = 1;
		for (int i = 1; i <= n; i++) {
			b[i] = b[i - 1] + (a[i] == '1');
			int pos = 2 * b[i] - i + n + 1;
			int x = c[pos];
			ans = (ans + 1ll * x * (n - i + 1) % mod) % mod;
			while (ans < 0) ans += mod;
			c[pos] = (c[pos] + 1 + i) % mod;
		}
		for (int i = 0; i <= 2 * n + 1; i++) c[i] = 0;
		writeln(ans);
	}
//	fwrite(pf, 1, o1 - pf, stdout);
	return 0;
}

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F. Bomb

由单调性考虑二分来求出 $k$ 次操作是否能把所有数减到小于等于某个 $x$， 再由这个边界去计算答案。

const int N = 2e5 + 5;
const int mod = 1e9 + 7;
#define int long long
int T, n, k, a[N], b[N];

inline bool check(int x) {
	int cnt = 0;
	for (int i = 1; i <= n; i++) if (a[i] >= x) {
		cnt += (a[i] - x) / b[i] + 1;
	}
	return cnt >= k;
}

inline ll calc(int x) {
	ll ret = 0; int cnt = 0;
	for (int i = 1; i <= n; i++) if (a[i] >= x) {
		int t = (a[i] - x) / b[i] + 1;
		cnt += t;
		ret += 1ll * t * a[i] - 1ll * t * (t - 1) / 2ll * b[i];
	}
	if (x > 0) ret += 1ll * (k - cnt) * x;
	return ret;
}

signed main(void) {
	for (read(T); T; T--) {
		read(n), read(k); int x = 0;
		for (int i = 1; i <= n; i++) read(a[i]), chkmax(x, a[i]);
		for (int i = 1; i <= n; i++) read(b[i]);
		int l = 1, r = x, ans = 0;
		while (l <= r) {
			int mid = l + r >> 1;
			if (check(mid)) ans = mid, l = mid + 1;
			else r = mid - 1;
		}
		writeln(calc(ans));
	}
//	fwrite(pf, 1, o1 - pf, stdout);
	return 0;
}

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G. Penacony

两种做法：一个是常规的线段树扫描线，因为确定一种取法后，按顺序枚举删去的边，所有路径选法都确定了，再加上只要按顺序每条路径方向只变化两次，可以线段树维护区间加减和区间 $0$ 的个数来做。另一种做法是 $Xor-Hash$ ,我们发现，选择修改某一条路径的方向，就是翻转了全局对于这条路径的选择情况，容易想到差分异或。给每对点一个随机值，再求前缀。只要维护出现次数最多的数即可（将他们统统去除，维护他们的补集）。

mt19937_64 rnd((unsigned int)chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 5;
int T, n, m, mx; ll a[N];

unordered_map <ll, int> h;

signed main(void) {
	for (read(T); T; T--) {
		read(n), read(m); h.clear(); mx = 0;
		for (int i = 1; i <= n; i++) a[i] = 0;
		while (m--) {
			int x, y; ll r = rnd();
			read(x), read(y);
			a[x] ^= r; a[y] ^= r;
		}
		for (int i = 1; i <= n; i++) a[i] ^= a[i - 1], h[a[i]]++;
		for (auto u : h) chkmax(mx, u.second);
		writeln(n - mx);
	}
	//fwrite(pf, 1, o1 - pf, stdout);
	return 0;
}