D-小杜捕鱼_周赛 (nowcoder.com)

如果使用二分会超时

#include<bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int n, m, ans;
char a[N][N];
bool check(int x, int aa, int bb)
{for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){if(a[i][j] == '#'){if(abs(aa - i) + abs(bb - j) >= x)return false;}}}return true;
}
int main()
{cin >> n >> m;for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){cin >> a[i][j];}}for(int i = 1; i <= n; i ++){for(int j = 1; j <= m; j ++){int l = 0, r = 1e9 + 10;while(l + 1 != r){int mid = (l + r) >> 1;if(check(mid, i, j))r = mid;else l = mid;}ans = max(ans, l);}}cout << ans;return 0;
}

故我们可以进行一定的优化

对于每条鱼我们知道，四个顶点就是可以到达的最远的距离

#include <bits/stdc++.h>
using namespace std;const int N = 1e3 + 5, INF = 0x3f3f3f3f, MOD = 1e9 + 7;#define x first;
#define y second;typedef long long LL;
typedef pair<int, int> PII;int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};int n, m;
char g[N][N];int main()
{ios::sync_with_stdio(false);cin.tie(0);cin >> n >> m;for (int i = 1; i <= n; i ++)for (int j = 1; j <= m; j ++)cin >> g[i][j];int ans = -1;for (int i = 1; i <= n; i ++)for (int j = 1; j <= m; j ++)if (g[i][j] == '#'){int d1 = i - 1 + j - 1; 			int d = d1;int d2 = i - 1 + abs(j - m);		d = max(d, d2);int d3 = abs(i - n) + j - 1;		d = max(d, d3);int d4 = abs(i - n) + abs(j - m);	d = max(d, d4);ans = max(ans, d);// cout << d1 << ' ' << d2 << ' ' << d3 << ' ' << d4 << endl;}cout << ans << endl;return 0;
}

F-小红的回文串_周赛 (nowcoder.com)

回文串可以对半遍历，如果遍历到的为’?'则看对应的是否为'?'，如果是则说明这里可以有26种变化，如果不是则说明只有一种变化就为对应的字母

注：如果字符串是奇数，则特别注意中间的字母如果中间的字母为’?'则又有了26种变化

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
string s;
int sum;
int main()
{cin >> s;int n = s.size();for(int i = 0; i < n / 2; i ++){if(s[i] == '?' && s[n - i - 1] == '?')sum ++;if(s[i] != s[n - i - 1]){if(s[i] != '?' && s[n - i - 1] != '?'){cout << 0;return 0;}}}long long ans = 1;if(n & 1 && s[n / 2] == '?')sum ++;for(int i = 0; i < sum; i ++)ans = (ans * 26) % mod;cout << ans;return 0;
}