「蓝桥」0208 第 26 场 蓝桥入门赛

Author: gdut-yy

lanqiao/src/main/java/lq250208/LQ250208T1.cpp

@@ -0,0 +1,18 @@
+import java.util.Arrays;
+import java.util.Scanner;
+
+public class a {
+ public static void main(String[] args) {
+ Scanner scanner = new Scanner(System.in);
+ System.out.println(solve());
+ }
+
+ private static String solve() {
+ char[] s = "snake".toCharArray();
+ Arrays.sort(s);
+ return String.valueOf(s);
+ }
+}
+/*
+蛇年大吉【算法赛】
+ */

lanqiao/src/main/java/lq250208/LQ250208T2.cpp

@@ -0,0 +1,35 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef long long ll;
+
+void solve() {
+ int n;
+ string a, b;
+ cin >> n >> a >> b;
+
+ // 统计字符串 a 中 0 和 1 的数量
+ int x = std::count(a.begin(), a.end(), '0');
+ int y = n - x;
+
+ // 统计字符串 b 中 0 和 1 的数量
+ int d = std::count(b.begin(), b.end(), '0');
+ int e = n - d;
+
+ // 计算最大异或值中 1 的个数
+ int ans = min(x, e) + min(d, y);
+ cout << ans << endl;
+}
+
+signed main() {
+ ios::sync_with_stdio(false);
+ cin.tie(nullptr);
+
+ int t = 1;
+// cin >> t;
+ while (t--) solve();
+ return 0;
+}
+/*
+对联【算法赛】
+ */

lanqiao/src/main/java/lq250208/LQ250208T2.java

@@ -0,0 +1,40 @@
+package lq250208;
+
+import java.util.Arrays;
+import java.util.Scanner;
+
+public class LQ250208T2 {
+ static int n;
+ static char[] a, b;
+
+ public static void main(String[] args) {
+ Scanner scanner = new Scanner(System.in);
+ n = scanner.nextInt();
+ a = scanner.next().toCharArray();
+ b = scanner.next().toCharArray();
+ System.out.println(solve());
+ }
+
+ private static String solve() {
+ Arrays.sort(a);
+ reverseSort(b);
+
+ int ans = 0;
+ for (int i = 0; i < n; i++) {
+ ans += a[i] ^ b[i];
+ }
+ return String.valueOf(ans);
+ }
+
+ static void reverseSort(char[] arr) {
+ Arrays.sort(arr);
+ for (int l = 0, r = arr.length - 1; l < r; l++, r--) {
+ char tmp = arr[l];
+ arr[l] = arr[r];
+ arr[r] = tmp;
+ }
+ }
+}
+/*
+对联【算法赛】
+ */

lanqiao/src/main/java/lq250208/LQ250208T3.cpp

@@ -0,0 +1,40 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef long long ll;
+
+void solve() {
+ string s;
+ cin >> s;
+
+ int x = 0, y = 0;
+ for (const auto &b: s) {
+ if (b == 'U') y++;
+ if (b == 'D') y--;
+ if (b == 'L') x++;
+ if (b == 'R') x--;
+ }
+ if (s.size() % 2 == 1) {
+ cout << -1 << endl;
+ } else {
+ int ans = min(abs(x), abs(y));
+ ans += abs(abs(x) - abs(y)) / 2;
+ cout << ans << endl;
+ }
+}
+
+signed main() {
+ ios::sync_with_stdio(false);
+ cin.tie(nullptr);
+
+ int t = 1;
+// cin >> t;
+ while (t--) solve();
+ return 0;
+}
+/*
+电子舞龙【算法赛】
+
+相似题目: 3443. K 次修改后的最大曼哈顿距离
+https://leetcode.cn/problems/maximum-manhattan-distance-after-k-changes/description/
+ */

lanqiao/src/main/java/lq250208/LQ250208T4.cpp

@@ -0,0 +1,41 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef long long ll;
+
+void solve() {
+ int n;
+ cin >> n;
+
+ vector<int> v(n);
+ for (int i = 0; i < n; ++i) {
+ cin >> v[i];
+ }
+ std::sort(v.begin(), v.end());
+
+ priority_queue<int, vector<int>, greater<int>> pq;
+ pq.push(0);
+ for (int i = 0; i < n; ++i) {
+ int val = pq.top();
+ if (val < v[i]) {
+ pq.pop();
+ pq.push(val + 1);
+ } else {
+ pq.push(1);
+ }
+ }
+ cout << pq.size() << endl;
+}
+
+signed main() {
+ ios::sync_with_stdio(false);
+ cin.tie(nullptr);
+
+ int t = 1;
+// cin >> t;
+ while (t--) solve();
+ return 0;
+}
+/*
+舞狮【算法赛】
+ */

lanqiao/src/main/java/lq250208/LQ250208T5.cpp

@@ -0,0 +1,35 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef long long ll;
+
+void solve() {
+ int n, k;
+ cin >> n >> k;
+
+ vector<int> a(n);
+ for (int i = 0; i < n; ++i) {
+ cin >> a[i];
+ a[i] %= k;
+ }
+
+ std::sort(a.begin(), a.end());
+ int ans = a[n - 1] - a[0];
+ for (int i = 0; i < n - 1; ++i) {
+ ans = min(ans, a[i] + k - a[i + 1]);
+ }
+ cout << ans << endl;
+}
+
+signed main() {
+ ios::sync_with_stdio(false);
+ cin.tie(nullptr);
+
+ int t = 1;
+// cin >> t;
+ while (t--) solve();
+ return 0;
+}
+/*
+扑克较量【算法赛】
+ */

lanqiao/src/main/java/lq250208/LQ250208T6.cpp

@@ -0,0 +1,41 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef long long ll;
+const int mod = 1e9 + 7;
+
+ll ksm(ll x, ll k, ll mo = mod) {
+ ll ans = 1;
+ while (k) {
+ if (k & 1) ans = ans * x % mo;
+ x = x * x % mo;
+ k >>= 1;
+ }
+ return ans;
+}
+
+void solve() {
+ ll a, b, c, n;
+ cin >> a >> b >> c >> n;
+
+ ll d = a * b % mod * c % mod;
+ ll ans = ksm(d, ksm(2, n, mod - 1), mod);
+ cout << ans << endl;
+}
+
+signed main() {
+ ios::sync_with_stdio(false);
+ cin.tie(nullptr);
+
+ int t = 1;
+ cin >> t;
+ while (t--) solve();
+ return 0;
+}
+/*
+春晚魔术【算法赛】
+
+可以归纳出，经过 N 次变换后，乘积 P_N = P_{0}^{2^N}。
+因此，我们只需要计算结果初始乘积 P0，然后计算 P0 的 2^N 次方即可。
+需要注意的是，2N 可能很大，因此我们需要使用快速幂并搭配欧拉降幂来完成计算。
+ */