Codeforces Round 908 (Div. 2)

Codeforces Round 908 (Div. 2)

A

获取对战的最后一个即为胜者

#include <bits/stdc++.h>using namespace std;void solve()
{int n;cin >> n;string s;cin >> s;cout << s[n-1] << endl;
}int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin >> T;while(T --){solve();}return 0;
}

B

只要有两个出现两次以上的数字即为合法
对着两个数字分别使用两种组合即可

#include <bits/stdc++.h>using namespace std;
const int N = 1e5;int a[N];void solve()
{int n;unordered_map<int, int> m;cin >> n;for (int i = 1; i <= n; i++){cin >> a[i];m[a[i]]++;}int s1 = -1, s2 = -1, cnt = 0;for (auto &&[x, y] : m){if (y >= 2){if (s1 == -1)s1 = x;elses2 = x;cnt++;if (cnt == 2)break;}}if (cnt < 2){cout << "-1\n";return;}vector<int> k(5);for (int i = 1; i <= n; i++){if (a[i] == s1){if (k[1] == 0){cout << "1 ";k[1] = 1;}else if (k[2] == 0){cout << "2 ";k[2] = 1;}elsecout << "1 ";}else if (a[i] == s2){if (k[3] == 0){cout << "1 ";k[3] = 1;}else if (k[4] == 0){cout << "3 ";k[4] = 1;}elsecout << "1 ";}else{cout << "1 ";}}cout << endl;
}int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin >> T;while (T--){solve();}return 0;
}

C

执行操作后， x总是数组的最后一个元素。
若x大于n则非法
旋转操作至多进行min(n,k)次

#include <bits/stdc++.h>using namespace std;const int N = 2e5 + 10;void solve()
{int n, k;cin >> n >>  k;vector<int>a(n);for (int i = 0; i < n; i++)cin >> a[i];int last = n - 1;for (int i = 0; i < min(k,n); i++){if (a[last] > n){cout << "No\n";return;}last += n - a[last];if (last >= n)last -= n;}cout << "Yes\n";
}int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin >> T;while (T--){solve();}return 0;
}

D

b数组顶多贡献一次反向排序再归并就是最小的方案

#include <bits/stdc++.h>using namespace std;const int N = 2e5 + 10;void solve()
{int n , m;cin >> n >> m;vector<int> a(n), b(m), c(n + m);for(int i = 0 ; i < n ; i ++)cin >> a[i];for(int i = 0 ; i < m ; i ++)cin >> b[i];sort(b.rbegin() , b.rend()); //反向排序merge(a.begin(), a.end(), b.begin(), b.end(),c.begin(), greater<int>()); //归并排序for (int i = 0; i < n + m; i++) {cout << c[i] << ' ';}cout << '\n';
}int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin >> T;while (T--){solve();}return 0;
}

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