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Commit 233c416

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「蓝桥」0208 第 26 场 蓝桥入门赛
1 parent 3f1cdea commit 233c416

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7 files changed

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7 files changed

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import java.util.Arrays;
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import java.util.Scanner;
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public class a {
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in);
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System.out.println(solve());
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}
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private static String solve() {
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char[] s = "snake".toCharArray();
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Arrays.sort(s);
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return String.valueOf(s);
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}
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}
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/*
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蛇年大吉【算法赛】
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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void solve() {
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int n;
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string a, b;
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cin >> n >> a >> b;
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// 统计字符串 a 中 0 和 1 的数量
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int x = std::count(a.begin(), a.end(), '0');
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int y = n - x;
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// 统计字符串 b 中 0 和 1 的数量
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int d = std::count(b.begin(), b.end(), '0');
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int e = n - d;
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// 计算最大异或值中 1 的个数
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int ans = min(x, e) + min(d, y);
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cout << ans << endl;
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}
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signed main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int t = 1;
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// cin >> t;
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while (t--) solve();
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return 0;
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}
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/*
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对联【算法赛】
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*/
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package lq250208;
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import java.util.Arrays;
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import java.util.Scanner;
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public class LQ250208T2 {
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static int n;
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static char[] a, b;
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in);
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n = scanner.nextInt();
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a = scanner.next().toCharArray();
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b = scanner.next().toCharArray();
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System.out.println(solve());
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}
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private static String solve() {
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Arrays.sort(a);
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reverseSort(b);
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int ans = 0;
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for (int i = 0; i < n; i++) {
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ans += a[i] ^ b[i];
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}
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return String.valueOf(ans);
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}
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static void reverseSort(char[] arr) {
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Arrays.sort(arr);
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for (int l = 0, r = arr.length - 1; l < r; l++, r--) {
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char tmp = arr[l];
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arr[l] = arr[r];
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arr[r] = tmp;
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}
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}
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}
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/*
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对联【算法赛】
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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void solve() {
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string s;
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cin >> s;
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int x = 0, y = 0;
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for (const auto &b: s) {
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if (b == 'U') y++;
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if (b == 'D') y--;
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if (b == 'L') x++;
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if (b == 'R') x--;
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}
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if (s.size() % 2 == 1) {
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cout << -1 << endl;
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} else {
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int ans = min(abs(x), abs(y));
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ans += abs(abs(x) - abs(y)) / 2;
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cout << ans << endl;
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}
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}
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signed main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int t = 1;
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// cin >> t;
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while (t--) solve();
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return 0;
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}
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/*
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电子舞龙【算法赛】
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相似题目: 3443. K 次修改后的最大曼哈顿距离
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https://leetcode.cn/problems/maximum-manhattan-distance-after-k-changes/description/
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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void solve() {
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int n;
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cin >> n;
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vector<int> v(n);
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for (int i = 0; i < n; ++i) {
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cin >> v[i];
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}
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std::sort(v.begin(), v.end());
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priority_queue<int, vector<int>, greater<int>> pq;
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pq.push(0);
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for (int i = 0; i < n; ++i) {
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int val = pq.top();
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if (val < v[i]) {
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pq.pop();
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pq.push(val + 1);
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} else {
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pq.push(1);
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}
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}
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cout << pq.size() << endl;
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}
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signed main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int t = 1;
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// cin >> t;
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while (t--) solve();
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return 0;
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}
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/*
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舞狮【算法赛】
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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void solve() {
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int n, k;
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cin >> n >> k;
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vector<int> a(n);
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for (int i = 0; i < n; ++i) {
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cin >> a[i];
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a[i] %= k;
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}
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std::sort(a.begin(), a.end());
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int ans = a[n - 1] - a[0];
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for (int i = 0; i < n - 1; ++i) {
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ans = min(ans, a[i] + k - a[i + 1]);
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}
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cout << ans << endl;
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}
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signed main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int t = 1;
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// cin >> t;
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while (t--) solve();
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return 0;
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}
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/*
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扑克较量【算法赛】
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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const int mod = 1e9 + 7;
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ll ksm(ll x, ll k, ll mo = mod) {
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ll ans = 1;
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while (k) {
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if (k & 1) ans = ans * x % mo;
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x = x * x % mo;
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k >>= 1;
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}
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return ans;
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}
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void solve() {
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ll a, b, c, n;
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cin >> a >> b >> c >> n;
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ll d = a * b % mod * c % mod;
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ll ans = ksm(d, ksm(2, n, mod - 1), mod);
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cout << ans << endl;
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}
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signed main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int t = 1;
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cin >> t;
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while (t--) solve();
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return 0;
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}
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/*
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春晚魔术【算法赛】
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可以归纳出,经过 N 次变换后,乘积 P_N = P_{0}^{2^N}。
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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。
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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。
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*/

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