「解题报告」freee Programming Contest 2023（AtCoder Beginner Contest 310）

比赛地址：freee Programming Contest 2023（AtCoder Beginner Contest 310） - AtCoder

后记：原本写了比较详细的题解，但是，突发意外情况，它没了，所以这份题解略写了。归根结底还是懒

A - Order Something Else#

贪心的选价格最小的和优惠券一起使用，最后和原价比大小即可。

/*
  The code was written by yifan, and yifan is neutral!!!
 */

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

template<typename T>
inline T read() {
	T x = 0;
	bool fg = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		fg |= (ch == '-');
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + (ch ^ 48);
		ch = getchar();
	}
	return fg ? ~x + 1 : x;
}

const int N = 110;

int n, q, p, mn = 1e9;
int d[N];

int main() {
	n = read<int>(), p = read<int>(), q = read<int>();
	for (int i = 1, x; i <= n; ++ i) {
		x = read<int>();
		mn = min(x, mn);
	}
	cout << min(p, q + mn) << '\n';
	return 0;
}

B - Strictly Superior#

B - Strictly Superior (atcoder.jp)

按照题意模拟即可。

/*
  The code was written by yifan, and yifan is neutral!!!
 */

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

template<typename T>
inline T read() {
	T x = 0;
	bool fg = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		fg |= (ch == '-');
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + (ch ^ 48);
		ch = getchar();
	}
	return fg ? ~x + 1 : x;
}

const int N = 110;

int n, m;
int f[N][N], p[N], c[N];
bitset<N> vis[N];

int main() {
	n = read<int>(), m = read<int>();
	for (int i = 1; i <= n; ++ i) {
		p[i] = read<int>(), c[i] = read<int>();
		for (int j = 1; j <= c[i]; ++ j) {
			vis[i].set(read<int>());
		}
	}
	for (int i = 1; i <= n; ++ i) {
		for (int j = 1; j <= n; ++ j) {
			if (i == j)	continue ;
			if (p[i] == p[j]) {
				if (((vis[i] & vis[j]) == vis[i]) && (c[j] > c[i])) {
					puts("Yes");
					return 0;
				}
			}
			else if (p[i] > p[j]) {
				if ((vis[i] & vis[j]) == vis[i]) {
					puts("Yes");
					return 0;
				}
			}
		}
	}
	puts("No");
	return 0;
}

C - Reversible#

字符串哈希来判断，可以搭配 map 来食用。

/*
  The code was written by yifan, and yifan is neutral!!!
 */

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

template<typename T>
inline T read() {
	T x = 0;
	bool fg = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		fg |= (ch == '-');
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + (ch ^ 48);
		ch = getchar();
	}
	return fg ? ~x + 1 : x;
}

const int N = 2e5 + 5;
const int based = 29;

int n, len, ans;
ull Hash1[N], Hash2[N];
char s[N];
map<ull, bool> mp;

int main() {
	n = read<int>();
	for (int i = 1; i <= n; ++ i) {
		scanf("%s", s + 1);
		len = strlen(s + 1);
		Hash1[0] = 0, Hash2[len + 1] = 0;
		for (int j = 1; j <= len; ++ j) {
			Hash1[j] = Hash1[j - 1] * based + s[j];
		}
		for (int j = len; j >= 1; -- j) {
			Hash2[j] = Hash2[j + 1] * based + s[j];
		}
		if (mp.count(Hash1[len]) || mp.count(Hash2[1])) {
			continue ;
		}
		mp[Hash1[len]] = 1;
		mp[Hash2[1]] = 1;
		++ ans;
	}
	cout << ans << '\n';
	return 0;
}

D - Peaceful Teams#

D - Peaceful Teams (atcoder.jp)

由于 $n$ 很小，所以可以直接搜索。

/*
  The code was written by yifan, and yifan is neutral!!!
 */

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

template<typename T>
inline T read() {
	T x = 0;
	bool fg = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		fg |= (ch == '-');
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + (ch ^ 48);
		ch = getchar();
	}
	return fg ? ~x + 1 : x;
}

int n, t, m, ans;
bool ok[15][15];
vector<int> Team[15];

void dfs(int u) {
	if (u == n + 1) {
		for (int i = 1; i <= t; ++ i) {
			if (!Team[i].size()) {
				return ;
			}
		}
		++ ans;
		return ;
	}
	for (int i = 1; i <= t; ++ i) {
		int fg = 0;
		for (int j : Team[i]) {
			fg |= ok[u][j];
		}
		if (fg) {
			continue ;
		}
		Team[i].emplace_back(u);
		dfs(u + 1);
		Team[i].pop_back();
		if (!Team[i].size()) {
			return ;
		}
	}
}

int main() {
	n = read<int>(), t = read<int>(), m = read<int>();
	for (int i = 1, x, y; i <= m; ++ i) {
		x = read<int>(), y = read<int>();
		ok[x][y] = ok[y][x] = 1;
	}
	dfs(1);
	cout << ans << '\n';
	return 0;
}

E - NAND repeatedly#

E - NAND repeatedly (atcoder.jp)

假设 $i$ 位置是 $1$，则到 $i - 1$ 位置的值，为 $1$ 的全变成 $0$，为 $0$ 的全变成 $1$，即交换即可，当 $i$ 位置是 $0$，那么所有情况都会变成 $1$，最后统计答案即可。

/*
  The code was written by yifan, and yifan is neutral!!!
 */

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

template<typename T>
inline T read() {
	T x = 0;
	bool fg = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		fg |= (ch == '-');
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + (ch ^ 48);
		ch = getchar();
	}
	return fg ? ~x + 1 : x;
}

const int N = 1e6 + 5;

int n;
ll ans;
string s;
ll a[N], cnt[2];

int main() {
	n = read<int>();
	for (int i = 1; i <= n; ++ i) {
		scanf("%1lld", &a[i]);
	}
	for (int i = 1; i <= n; ++ i) {
		if (a[i]) {
			swap(cnt[1], cnt[0]);
		}
		else {
			cnt[1] += cnt[0];
			cnt[0] = 0;
		}
		++ cnt[a[i]];
		ans += cnt[1];
	}
	cout << ans << '\n';
	return 0;
}

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F 往后没看题，就不写了。