C - Avoid K Palindrome 2 (atcoder.jp)
思路：
罗列出排列的每一种情况，再根据题目要求进行判断
代码：
void solve()
{ll n, k;cin >> n >> k;string s;vector<char>a;cin >> s;for (int i = 0; i < n; i++)a.push_back(s[i]);sort(a.begin(), a.end());bool ok, flag;ll ans = 0;while (true){ok = true;for (int i = 0; i <= n - k; i++){flag = true;for (int j = 0; j < k; j++){if (a[i + j] != a[i + 1 + k - j]){flag = false;}}if (flag)ok = false;}if (ok)ans++;if(!next_permutation(a.begin(),a.end()))break;}cout << ans << endl;return;
}D - Palindromic Number (atcoder.jp)
思路其实很简单，根据需要的i-th进行判断即可
#define _CRT_SECURE_NO_WARNINGS 1
//------ 棘手大学 世界第一 ------//
#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
//gcd最大公约数，lcm最小公倍数
typedef long long ll;
#define IOS ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
priority_queue<int, vector<int>, greater<int>> pq;
map<int, int>mp;
const int N = 2e6 + 10;
long long sum[60], a[60];
long long n;
int ans[1010], len, b[1010];
ll pow(int x) {long long res = 1;for (int i = 1; i <= x; i++)res *= 10;return res;
}
int main() {IOS;for (int i = 1; i <= 37; i++) {int mi = ((i + 1) / 2 - 1);a[i] = 1LL * 9 * pow(mi);sum[i] = sum[i - 1] + a[i];}cin >> n;if (n <= 1){cout << "0" << endl;return 0;}n -= 1;int pos = lower_bound(sum + 1, sum + 38, n) - sum;if (n == sum[pos - 1])pos--;while (n) {b[++len] = n % 10;n /= 10;}for (int i = 1, j = len; i <= len; i++, j--)ans[i] = b[j];ans[1]--; ans[len]++;if (ans[1] == 0 && ans[2] == 0)ans[2] = 9;int now = len;while (ans[now] >= 10) {ans[now - 1] += ans[now] / 10;ans[now] %= 10;now--;}int l = len, r = len + 1;if (pos % 2 == 1)l--;while (l >= 1) {ans[r] = ans[l];l--; r++;}l = 1, r--;while (ans[l] == 0)l++;while (ans[r] == 0)r--;for (int i = l; i <= r; i++)cout << ans[i];
}Problem - A - Codeforces
思路：
由题意可知两者每次肯定都会做出最佳选择（即选取最大），所以我们只需要判断最大值是奇数还是偶数即可（因为是Alice先手，所以奇数A必胜，否则B胜）
代码：
//该段代码经过修改仅提供思路，不能AC
void solve()
{ll a[N];ll n;cin >> n;for (int i = 1; i <= n; i++){cin >> a[n];mp[a[n]]++;}sort(a + 1, a + 1 + n, cmp);ll flag = 1;for (auto k : a){if (mp[a[k]] % 2 != 0){flag = 0;}}if (flag)cout << "NO" << endl;elsecout << "NO" << endl;return;
}Problem - B - Codeforces
代码：
#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
#include <stdbool.h>#define N 2000010
typedef long long ll;ll n, t, x, y;
ll dp[N];void solve() {scanf("%lld %lld %lld", &n, &x, &y);for (ll i = 1; i <= n; i++) {if (i >= y && i <= x) {dp[i] = 1;}}for (ll i = y - 1; i >= 1; i -= 2) {dp[i] = -1;if (i - 1 >= 1) {dp[i - 1] = 1;}}for (ll i = x + 1; i <= n; i += 2) {dp[i] = -1;if (i + 1 <= n) {dp[i + 1] = 1;}}for (ll i = 1; i <= n; i++)printf("%lld ", dp[i]);printf("\n");
}int main() {scanf("%lld", &t);while (t--) {solve();}return 0;
}Problem - C - Codeforces
代码：
void solve() {int n;cin >> n;vector<ll> a(n);for (int i = 0; i < n; ++i)cin >> a[i];ll res = 0;for (int i = 0; i < 2; ++i) {vector<bool> vis(n + 1);ll ma = 0;for (ll& x : a) {if (vis[x])ma = max(ma, x);vis[x] = true;res += x;x = ma;}}for (int i = 0; i < n; ++i)res += (n - i) * a[i];cout << res << endl;
}C-嘤嘤不想买东西喵_牛客周赛 Round 49(重现赛) (nowcoder.com)
本质就是求最大连续子段和
代码:
#define _CRT_SECURE_NO_WARNINGS 1
//------ 棘手大学 世界第一 ------//
#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
//gcd最大公约数，lcm最小公倍数
typedef long long ll;
#define IOS ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
priority_queue<int, vector<int>, greater<int>> pq;
map<int, int>mp;
const int N = 2e6 + 10;
int mod(string a, ll b)//高精度a除以单精度b
{ll n = 0;for (int i = 0; i < a.size(); i++) n = (n * 10 + (a[i] - '0')) % b; //求出余数return n;
}
string gcd(string a, ll b) // 欧几里得算法
{while (b != 0){ll temp = b;b = mod(a, b);a = to_string(temp);}return a;
}
int gcd1(int a, int b)//a、b不可以为0(很快)
{while (b ^= a ^= b ^= a %= b);return a;
}
int gcd2(int a, int b)//a、b可以为0(很快)
{if (b) while ((a %= b) && (b %= a));return a + b;
}
int lcm1(int a, int b) {return a * b / gcd1(a, b);
}
bool cmp(ll x, ll y)
{return x > y;
}
ll a[N], dp[N];
void solve()
{ll n, x;cin >> n >> x;for (int i = 1; i <= n; i++)cin >> a[i], a[i] -= x;ll ans = 0;for (int i = 1; i <= n; i++){dp[i] = max(a[i], dp[i - 1] + a[i]);}for (int i = 1; i <= n; i++){ans = max(ans, dp[i]);}cout << ans << endl;return;
}
int main()
{IOS;solve();return 0;
}
D-嘤嘤不想求异或喵_牛客周赛 Round 49(重现赛) (nowcoder.com)
求区间异或和（以后直接套用模板算了）
代码：
ll xor_prefix(ll n) {switch (n % 4) {case 0: return n;case 1: return 1;case 2: return n + 1;case 3: return 0;}return 0;
}
int xor_range(ll L, ll R) {//区间所有整数异或和return xor_prefix(R) ^ xor_prefix(L - 1);
}
