Codeforces Round 1116 (Div. 2)

发表于 本文字数： 879 阅读时长 ≈ 3 分钟

A

下界始终是三个数中的最小值，上界要么是最大值，要么是除最大值外的两个数之和

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"
// #define int long long
const int mod = 998244353;
const int maxn = 2e5 + 5;
int a[maxn];
void Solve() {
	for (int i = 1; i <= 3; i++) cin >> a[i];
	sort(a + 1, a + 1 + 3);
	cout << min(a[3] - a[1], a[2]) << endl;
}
signed main() {
    // freopen("a.in", "r", stdin);
    // freopen("a.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        Solve();
    }
    return  0;
}

B

观察发现合法串一定是 0000 和 1111 交替。因此只需要枚举四种开头 01100110, 00110011, 11001100, 10011001。

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"
// #define int long long
const int mod = 998244353;
const int maxn = 2e5 + 5;
int n;
string s;
void Solve() {
	cin >> n >> s;
	int ans = 0;
	++ans;
	for (int i = 0; i < n; i++) {
		if ((i % 4 == 0 || i % 4 == 1)) {
			if (s[i] == '1') {
				--ans; break;
			}
		} else {
			if (s[i] == '0') {
				--ans; break;
			}
		}
	}
	++ans;
	for (int i = 0; i < n; i++) {
		if (i % 4 == 0 || i % 4 == 3) {
			if (s[i] == '1') {
				--ans; break;
			}
		} else {
			if (s[i] == '0') {
				--ans; break;
			}
		}
	}
	++ans;
	for (int i = 0; i < n; i++) {
		if (i % 4 == 2 || i % 4 == 3) {
			if (s[i] == '1') {
				--ans; break;
			}
		} else {
			if (s[i] == '0') {
				--ans; break;
			}
		}
	}
	++ans;
	for (int i = 0; i < n; i++) {
		if (i % 4 ==1 || i % 4 == 2) {
			if (s[i] == '1') {
				--ans; break;
			}
		} else {
			if (s[i] == '0') {
				--ans; break;
			}
		}
	}
	cout << ans << endl;
}
signed main() {
    // freopen("b.in", "r", stdin);
    // freopen("b.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        Solve();
    }
    return  0;
}

C

显然，初始有土豆的人 应该在最后一轮的时候，将土豆传递给下一个人，这样就能保证自己手上的土豆最后一定落在对方手上。注意判断下一个人也有土豆的情况就行了。

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"
// #define int long long
const int mod = 998244353;
const int maxn = 2e5 + 5;
int n, k;
string s;
void Solve() {
	cin >> n >> k;
	cin >> s;
	int ans[2] = {0};
	for (int i = 0; i < n * 2; i++) {
		if (s[i] == '0') continue;
		if (s[(i + 1) % (n * 2)] == '1') ++ans[!(i % 2)];
		else ++ans[i % 2];
	}
	cout << ans[0] << " " << ans[1] << endl;
}
signed main() {
    // freopen("c.in", "r", stdin);
    // freopen("c.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        Solve();
    }
    return  0;
}

D

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"
#define int long long
const int mod = 998244353;
const int maxn = 1e6 + 5;
int frac[maxn];
int qpow(int a, int n) {
	int res = 1;
	while (n) {
		if (n & 1) res = 1ll * res * a % mod;
		a = 1ll * a * a % mod;
		n /= 2;
	}
	return res;
}
int C(int n, int m) {
	if (m < 0 || m > n || n < 0) return 0;
	return 1ll * frac[n] * qpow(1ll * frac[m] * frac[n - m] % mod, mod - 2) % mod;
}
int n;
string s;
void Solve() {
	cin >> n;
	cin >> s;
	int k = 0, t = 0;
	for (int i = 1; i < n; i++) {
		if (s[i - 1] != s[i]) {
			++k;
			if (k & 1) t += i;
			else t -= i;
		}
	}
	if (k == 0) {
		cout << 1 << endl;
		return;
	} else if (k & 1) {
		cout << C(t - 1, k / 2) * C(n - 1 - t, k / 2) % mod << endl;
	} else {
		cout << C(-t - 1, k / 2 - 1) * C(n - 1 + t, k / 2) % mod << endl;
	}
}
signed main() {
    // freopen("d.in", "r", stdin);
    // freopen("d.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int T;
    cin >> T;
	frac[0] = 1;
	for (int i = 1; i < maxn; i++) frac[i] = 1ll * frac[i - 1] * i % mod;
    while (T--) {
        Solve();
    }
    return  0;
}

F

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"
#define int long long
const int mod = 998244353;
const int maxn = 2e5 + 5;
int n;
int a[maxn];
bool check(int k) {
	priority_queue<int> q;
	for (int i = 1; i <= n; i++) q.push(a[i]);
	for (int i = 0; i < max(0ll, k - 30); i++) {
		q.pop(); q.push(0);
	}
	for (int i = min(k - 1, 29ll); i >= 0; i--) {
		int x = q.top();
		q.pop();
		x = max(0ll, x - (1ll << i));
		q.push(x);
	}
	return q.top() == 0;
}
void Solve() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	int l = 0, r = n + 30, ans = -1;
	while (l <= r) {
		int mid = (l + r) / 2;
		if (check(mid)) {
			r = mid - 1;
			ans = mid;
		} else {
			l = mid + 1;
		}
	}
	cout << ans << endl;
}
signed main() {
    // freopen("f.in", "r", stdin);
    // freopen("f.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        Solve();
    }
    return  0;
}