外卖网站建设方案书个人主页怎么申请

2025.2.17——1400

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A 1400

B 1400

C 1400

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• 二分+构造+字符串/贪心/思维。CF的题就得多看透几层表面发掘本质。

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A

1. 一眼单调性。
2. 分析后可以二分答案。

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B

1. 本质是：$(j-i)|n,s[i]!=s[j]$ 。设想一个周期为 $T$ 。
2. 设最小的不能被 $n$ 整除的数为 $t$ 。必要性：$T>=t$ 。充分性：$T==t$ 时，$s[i]==s[j]$的距离为 $t$ 的倍数，不能被 $n$ 整除，满足题意。
3. 绝妙的构造…

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C

1. 字符串贪心匹配。看着题解想了好久，值得多几次回味。解法和优化亦多样，正难则反/搜索/动态规划…

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------------------------代码------------------------

A

#include <bits/stdc++.h>
#define int long long
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC)               \{                           \for (auto Vec : VEC)    \cout << Vec << ' '; \cout << '\n';           \}void _();
signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cout << fixed << setprecision(10);int T = 1;cin >> T;while (T--)_();return 0;
}void _()
{int n;cin >> n;vector<int> a(n + 1);for (int i = 1; i <= n; i++)cin >> a[i];sort(begin(a) + 1, end(a));// bugv(a);auto ok = [&](int x){int cnt = 0;// bug2(1, x);for (int i = 1; i <= n;){int j = i;cnt++;for (; j <= n && a[j] - a[i] <= x << 1; j++);i = j;// bug(i);}// bug2(2, cnt);return cnt <= 3;};// ok(2);int l = -1, r = 1e9;while (r - l - 1){int mid = l + r >> 1;if (ok(mid))r = mid;elsel = mid;}cout << r << '\n';
}

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B

#include <bits/stdc++.h>
#define int long long
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC)               \{                           \for (auto Vec : VEC)    \cout << Vec << ' '; \cout << '\n';           \}void _();
signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cout << fixed << setprecision(10);int T = 1;cin >> T;while (T--)_();return 0;
}void _()
{int n;cin >> n;int st = 1;while (n % st == 0)st++;for (int i = 0; i < n; i++)cout << (char)('a' + i % st);cout << '\n';
}

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C

#include <bits/stdc++.h>
#define int long long
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
#define bugv(VEC)               \{                           \for (auto Vec : VEC)    \cout << Vec << ' '; \cout << '\n';           \}void _();
signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cout << fixed << setprecision(10);int T = 1;cin >> T;while (T--)_();return 0;
}void _()
{string s;cin >> s;int m;cin >> m;string l, r;cin >> l >> r;int st = 0;int n = s.size();s = ' ' + s;for (int i = 0; i < m; i++){int mx = st + 1;for (int j = l[i] - '0'; j <= r[i] - '0'; j++){int t = st + 1;while (t <= n && j != s[t] - '0')t++;mx = max(mx, t);}st = mx;}cout << (st > n ? "YES" : "NO") << '\n';
}

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