ABC 258 上分记录

挂了 $400$ 分

A

模拟，注意特判前导零。

signed main() 
{
	int x; cin >> x;
	int h = x / 60;
	x %= 60;
	printf("%d:", h + 21);
	if(x <= 10) printf("0");
	printf("%d",x);
	return 0;
}

B

我们发现他能够将这个矩形给复制成 $9$ 份。

然后在复制后的矩形里去找最大值就行了。

 
const int dx[] = {0, -1, -1, -1, 0, 0, 1, 1, 1};
const int dy[] = {0, -1, 0, 1, -1, 1, -1, 0, 1};
bitset <1000> vis[1000];
ll Ans = -0x3f3f3f3f;
 
void dfs(int x, int y, int fx, ll ans, int step) {	
	if(step == n + 1) return Ans = max(Ans, ans), void();
	int xx = x + dx[fx], yy = y + dy[fx];
	if(xx >= 1 && xx <= 3 * n && yy >= 1 && yy <= 3 * n) {
//		vis[xx][yy] = 1;
		dfs(xx, yy, fx, ans * 10 + mp[xx][yy], step + 1);
//		vis[xx][yy] = 0;
	}
}
 
signed main() 
{
	n = read();
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			scanf("%1d", &mp[i][j]);
		}
	}
	for(int i = 1; i <= n; i++) {
		for(int j = n + 1; j <= 2 * n; j++) {
			mp[i][j] = mp[i][j - n];
		}
	}
	for(int i = 1; i <= n; i++) {
		for(int j = 2 * n + 1; j <= 3 * n; j++) {
			mp[i][j] = mp[i][j - 2 * n];
		}
	}
	for(int i = n + 1; i <= 2 * n; i++) {
		for(int j = 1; j <= n; j++) {
			mp[i][j] = mp[i - n][j];
		}
	}
	for(int i = n + 1; i <= 2 * n; i++) {
		for(int j = n + 1; j <= 2 * n; j++) {
			mp[i][j] = mp[i][j - n];
		}
	}
	for(int i = n + 1; i <= 2 * n; i++) {
		for(int j = 2 * n + 1; j <= 3 * n; j++) {
			mp[i][j] = mp[i][j - 2 * n];
		}
	}
	for(int i = 2 * n + 1; i <= 3 * n; i++) {
		for(int j = 1; j <= n; j++) {
			mp[i][j] = mp[i - 2 * n][j];
		}
	}
	for(int i = 2 * n + 1; i <= 3 * n; i++) {
		for(int j =  n + 1; j <= 2 * n; j++) {
			mp[i][j] = mp[i][j - n];
		}
	}
	for(int i = 2 * n + 1; i <= 3 * n; i++) {
		for(int j = 2 * n + 1; j <= 3 * n; j++) {
			mp[i][j] = mp[i][j - 2 * n];
		}
	}
	for(int k = 1; k <= 8; k++) {
		for(int i = 1; i <= 3 * n; i++) {
			for(int j = 1; j <= 3 * n; j++) {
				dfs(i, j, k, 0, 1);
			}
		}
	}
//	printf("\n");
	cout << Ans;
	return 0;
}

C

sb 题，把字符串复制一遍，用一个指针来维护当前状态下的字符串。

其实可以对这个字符串建个主席树

signed main() {
	int n, Q;
	cin >> n >> Q;
	cin >> s;
	s += s;
	int l = 0;
	for (int i = 1; i <= Q; i++) {
		int opt = read(), x = read();
		if(opt & 1) l = (l + n - x) % n;
		}
		else cout << s[l + x - 1] << "\n";
	}
	return 0;
}

D

还是sb题，直接贪心就行了。

用一个前缀和还有前缀最小值来搞。

老套路了。

ll a[N], b[N];
ll ans = 2e18, qzh[N], c[N];
 
signed main() 
{
	int n = read(), x = read();
	c[0] = 1e18;
	for(int i = 1; i <= n; i++) {
		a[i] = read(), b[i] = read();
		qzh[i] = qzh[i - 1] + a[i] + b[i], c[i] = min(b[i] * 1ll, c[i - 1]);
	}
	for(int i = 1; i <= n; i++) {
		ll m = qzh[i];
		ans = min(ans, m + c[i] * (x - i));
	}
	cout << ans;
	return 0;
}

G

计数题。

暴力做法是 $\mathcal(O)(n^3)$

我们发现用一个 bitset 就做到了 $\mathcal(O)(\frac{n^3}{w})$

bitset <4000>  vis[4000];
 
signed main() 
{
	int n = read();
	for(int i = 1; i <= n; i++) {
		for(int j = n - 1; j >=  0; j--) {
			int x;
			scanf("%1d", &x);
			vis[i].set(j, x);
		}
	}
	ll ans = 0;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			if(vis[i][n - j]) ans += (vis[i] & vis[j]).count();
		}
	}
	cout << ans / 6;
	return 0;
}