Codeforces Round 940 (Div. 2) and CodeCraft-23

1|0A. Stickogon

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

#define int i64

using vi = vector<int>;

const int N = 3e5 , mod = 1e9+7; 

void solve(){
	vi cnt(101);
	int n;
	cin >> n;
	for(int i = 1, x; i <= n ; i ++)
		cin >> x, cnt[x] ++;
	int res = 0;
	for(int i = 1; i <= 100; i ++)
		res += cnt[i] / 3;
	cout << res << "\n";
}

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int TC;
	for(cin >> TC; TC; TC --)
		solve();
	return 0;
}

2|0B. A BIT of a Construction

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

#define int i64

using vi = vector<int>;

const int N = 3e5 , mod = 1e9+7; 

void solve(){
	int n, k;
	cin >> n >> k;
	if(n == 1) cout << k << "\n";
	else {
		int x = 0;
		while(x * 2 + 1 <= k) x = x * 2 + 1;
		cout << x << " " << k - x;
		for(int i = 1 ; i <= n - 2 ; i ++)
			cout << " 0";
		cout << "\n";
	}
}

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int TC;
	for(cin >> TC; TC; TC --)
		solve();
	return 0;
}

3|0C. How Does the Rook Move?

首先和格子的放置方式没有关系，只需要考虑剩下多少行没有被占。

$f[i]$ 表示有 $i$ 行没有被占领，则 $f[0] = f[1] = 1, f[i] = f[i-1] + 2\times C_{i-1}^{1} \times f[i-2]$

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;

#define int i64

using vi = vector<int>;

const int N = 3e5 , mod = 1e9+7; 

vi f(N+1);

void solve(){
	int n , m;
	cin >> n >> m;
	for(int x, y; m; m --) {
		cin >> x >> y;
		n -= 1 + (x != y);
	}
	cout << f[n] << "\n";
}

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	f[0] = f[1] = 1;
	for( int i = 2 ; i <= N; i ++ )
		f[i] = (f[i-1] + 2 * (i-1) * f[i-2]) % mod;
	int TC;
	for(cin >> TC; TC; TC --)
		solve();
	return 0;
}

4|0D. A BIT of an Inequality

题目的式子很容易化成 $f(x,z)\oplus a_y > f(x,z)$ 。

有一个性质是， $a_y$ 的最高位的 $1$ 如果是第 $h$ 位，则 $f(x,z)$ 第 $h$ 位必须是 $0$ ，其他位随意，所以直接统计哪些包含 $y$ 的区间满足条件，这里可以用到前缀和和后缀和实现。

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

#define int i64

using vi = vector<int>;

const int G = 30; 

void solve(){
	int n;
	cin >> n;
	vi a(n+1) , s(n + 1);
	for(int i = 1; i <= n; i ++) 
		cin >> a[i], s[i] = s[i - 1] ^ a[i];

	vector pre(G, vi(2));
	vector suf(n + 2, vector(G, vi(2)));
    
	for(int i = 0; i < G; i ++ )
		pre[i][0] = 1;
	for(int i = n; i >= 1 ; i -- ){
		suf[i] = suf[i+1];
		for(int j = 0; j < G; j ++)
			suf[i][j][(s[i] >> j) & 1] ++;
	}

	i64 res = 0;
	for(int i = 1, h; i <= n ; i ++){
		h = log2(a[i]);
		for(int l = 0; l < 2; l ++)
			res += pre[h][l] * suf[i][h][l];
		for(int j = 0; j < G; j ++)
			pre[j][(s[i] >> j) & 1] ++;
	}
	cout << res << "\n";
 	return ;
}

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int TC;
	for(cin >> TC; TC; TC --)
		solve();
	return 0;
}