Dashboard - 2023 CCPC Henan Provincial Collegiate Programming Contest - Codeforces

过题难度：A H F G B K E

铜奖： 2  339

银奖： 3  318

金奖： 5  523

A:

直接模拟

// Code Start Here	int t;cin >> t;while(t--){string s;cin >> s;set<char> seen;bool found = false;		for (int i = 1; i < sz(s); ++i) {bool ok = true;seen.clear();for (int j = 0; j < i; ++j) {if (seen.count(s[j])) {ok = false;break;}seen.insert(s[j]);}if (!ok) continue;int l = i, r = sz(s) - 1;while (l < r) {if (s[l] != s[r]) {ok = false;break;}++l;--r;}if (ok) {found = true;break;}}cout << (found ? "HE\n" : "NaN\n");}

H

思路：我们发现0.5会进位，而距离0.5无穷左边就会被舍去，因此最小的方式是分出来的尽量距离0.5差一点点，最多的就是0.5

// Code Start Here	int t;cin >> t;while(t--){int n , k;cin >> n >> k;if(k > 2 * n)cout << 0 << " " << 2 * n <<endl;else{int minv = (2 * n - k + 2) / 2;int maxv = minv + k - 1;cout << minv << " " << maxv << endl;}}

F

题意：

给定一个数列A从中选出 k 个数，计算：这些数的最大值-最小值，记作 max。这些数相邻差值中的最小值，记作 min。要求使max×min 最小

思路：我们发现排序后，任意 k 个数如果不连续，选中其中的最大和最小值之差一定不会比连续的更小。同时，相邻差值的最小值如果跳着选，差值可能会变大，而连续的差值是最稳定的。

对于连续的 k 个数，max=A[i+k−1]−A[i]。假设有k−1 个相邻差值，存在差分数组 diff，使得diff[j]=A[j+1]−A[j]。连续区间 [i, i+k-1] 中，相邻差值的最小值就是min=min(diff[i],diff[i+1],...,diff[i+k−2])。

因此我们可以遍历所有可能的连续 k 个数区间，求：每个区间内max×min，然后取最小值

考虑到5e5的数据规模，我们使用单调队列维护区间内差值的最小值。在O(1) 时间内取当前窗口最小值，且在移动窗口时自动维护单调性，效率是O(n)。

// Code Start Here	int n , k;cin >> n >> k;vector<int> a(n);for(int i = 0;i<n;i++)cin >> a[i];sort(all(a));vector<int> diff(n-1);for(int i = 0;i<n-1;i++)diff[i] = a[i+1] - a[i];int ans =  1e18;deque<int> dq;for(int i = 0;i<n;i++){if(i > 0){while(!dq.empty() && diff[i-1] <= diff[dq.back()])dq.pop_back();dq.push_back(i - 1);}if(i >= k){if (!dq.empty() && dq.front() < i - k)dq.pop_front();}if(i >= k-1){int max_v = a[i] - a[i-k+1];int min_v = 1e9;if(k > 1 && !dq.empty() )min_v = diff[dq.front()];if(k == 1)min_v = 1e9;ans = min(ans , 1LL *max_v * min_v);}}cout << ans << endl;return 0;

G：

一道超级大模拟题，考虑到个体的特殊性，我们可以对底数和幂数分别开两个数组来映射，使用的时候取出即可

char a[][100] = {"",".................................................................................",".................................................................................",".0000000.......1.2222222.3333333.4.....4.5555555.6666666.7777777.8888888.9999999.",".0.....0.......1.......2.......3.4.....4.5.......6.............7.8.....8.9.....9.",".0.....0.......1.......2.......3.4.....4.5.......6.............7.8.....8.9.....9.",".0.....0.......1.2222222.3333333.4444444.5555555.6666666.......7.8888888.9999999.",".0.....0.......1.2.............3.......4.......5.6.....6.......7.8.....8.......9.",".0.....0.......1.2.............3.......4.......5.6.....6.......7.8.....8.......9.",".0000000.......1.2222222.3333333.......4.5555555.6666666.......7.8888888.9999999.",".................................................................................",
};char b[][100] = {"",".............................................................",".00000.....1.22222.33333.4...4.55555.66666.77777.88888.99999.",".0...0.....1.....2.....3.4...4.5.....6.........7.8...8.9...9.",".0...0.....1.22222.33333.44444.55555.66666.....7.88888.99999.",".0...0.....1.2.........3.....4.....5.6...6.....7.8...8.....9.",".00000.....1.22222.33333.....4.55555.66666.....7.88888.99999.",".............................................................",".............................................................",".............................................................",".............................................................",
};char c[][100] = {"",".................................",".................................",".........IIIIIII.N.....N.FFFFFFF.","............I....NN....N.F.......",".=======....I....N.N...N.F.......","............I....N..N..N.FFFFFFF.",".=======....I....N...N.N.F.......","............I....N....NN.F.......",".........IIIIIII.N.....N.F.......",".................................",
};char ans[20][1000];
int tot;void add(ll x, int d){stack<int> q;if (x == 0) q.push(0);while (x) {q.push(x % 10);x /= 10;}while (!q.empty()) {int top = q.top();for (int i = 1; i <= 10; i++) {if (d == 1)for (int j = tot, k = top * 8; j < tot + 8; j++, k++)ans[i][j] = a[i][k];elsefor (int j = tot, k = top * 6; j < tot + 6; j++, k++)ans[i][j] = b[i][k];}if (d == 1) tot += 8;else tot += 6;q.pop();}
}void INV(int l, int r){for (int i = 1; i <= 10; i++)for (int j = tot, tl = l; tl <= r; j++, tl++)ans[i][j] = c[i][tl];tot = tot + (r - l + 1);
}bool check(ll x, ll y){if (x == 1) return false;ll a = 1e18;for (ll i = 1; i <= y; i++) {if (a < x) return true;a /= x;}return false;
}ll qp(ll x, ll y){if (x == 1) return 1;ll t = 1;for (; y; y--) {if (t > 1e18 / x) return 1e18 + 1;t = t * x;}return t;
}
int main(){int T;cin >> T;cin.ignore();for (int t = 0; t < T; t++) {string s;getline(cin, s);ll x = 0, y = 0;size_t pos = s.find("^{");string xs = s.substr(0, pos);string ys = s.substr(pos + 2, s.size() - pos - 3);x = stoll(xs);y = stoll(ys);tot = 0;add(x, 1);add(y, 2);INV(0, 7);if (check(x, y))INV(8, 31);elseadd(qp(x, y), 1);for (int i = 1; i <= 10; i++) {ans[i][tot] = 0;printf("%s.\n", ans[i]);}printf("\n");}
}

B

发现对于ai>aj，i<j，那么i和j必须在同一个块里。否则任意即可

所以可以直接划分成最小的不可再分的几个区间。从l到i，i是右端点当且仅当区间最大值小于等于后面的最小值。把这些右端点标记一下，满足答案的必须满足是倍数

预处理一下后缀，再暴力扫一遍倍数即可

// Code Start Here	int n;cin >> n;vector<int> a(n+10) , m(n+10) , f(n+10);for(int i = 1;i<=n;i++)cin >> a[i];m[n+1] = a[n];for(int i = n;i>=1;i--)m[i] = min(m[i+1],a[i]);int max_val = 0;for(int i = 1;i<=n;i++){max_val = max(max_val , a[i]);if(max_val <= m[i + 1] || i == n){f[i] = 1;max_val = 0;}}int ans = 0;for(int i = 1;i<=n;i++){ans ++;for(int j = i;j<=n;j+=i){if(f[j] == 0){ans--;break;}}}cout << ans <<endl;

K

题意：构造一个序列使得相邻两个数差的绝对值为质数

对于n < 10可以暴力枚举，对于n > 10

若n为奇数，可以1 3 5 ... n-2 .. n .. n-3 .. n-5 .. 6 4 2

若n为偶数，可以1 3 5 .. n-3  n n-2 ...6 4 2

// Code Start Here	
int n;cin >> n;
switch (n) {
case 1: case 2: case 3: case 4:cout << "-1";break;
case 5:cout << "5 3 1 4 2";break;
case 6:cout << "5 3 1 6 4 2";break;
case 7:cout << "6 3 5 2 7 4 1";break;
case 8:cout << "6 3 8 5 2 7 4 1";break;
case 9:cout << "1 3 8 5 7 2 9 6 4";break;
default:if (n % 2) {for (int i = 4; i <= n; i += 2) {cout << i << ' ';if (n - 7 == i) cout << n - 2 << ' ';if (n - 5 == i) cout << n << ' ';}for (int i = n - 4; i >= 1; i -= 2) {cout << i << ' ';if (i == 7) cout << "2 ";}} else {for (int i = 4; i <= n; i += 2) {cout << i << ' ';if (n - 6 == i) cout << n - 1 << ' ';}for (int i = n - 3; i >= 1; i -= 2) {cout << i << ' ';if (i == 7) cout << "2 ";}}
}

E

考虑到可以使用KMP算法来计算最大前后缀，我们跑一遍KMP然后对ne数组去一个最大值即可，当最大值大于 50就可以考虑是错的

// Code Start Here	string s;cin >> s;vector<int> ne(sz(s)+10 , 0);for(int i = 1;i<sz(s);i++){int j = ne[i - 1];while(j && s[i] != s[j]){j = ne[j - 1];}if(s[j] == s[i])j++;ne[i] = j;if(ne[i] >= 50){cout <<"No" << endl;return 0;}}cout <<"Yes" <<endl;