Codeforces Round 968 (Div. 2)

合集 - codeforces(10)

1.CodeTON Round 12024-05-172.Educational Codeforces Round 1652024-06-293.Codeforces Round 364 (Div. 2)2024-06-294.Codeforces Round 958 (Div. 2)2024-07-265.Codeforces Round 947 (Div. 1 + Div. 2)2024-08-136.EPIC Institute of Technology Round August 2024 (Div. 1 + Div. 2)2024-08-147.Codeforces Round 966 (Div. 3)2024-08-15

8.Codeforces Round 968 (Div. 2)2024-09-15

9.Codeforces Round 969 (Div. 2)2024-09-1610.Codeforces Round 976 (Div. 2)2024-09-30

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A.#

判断首位字符是否相等即可

#include<bits/stdc++.h>
using namespace std;

const int N = 1e6 + 7;
void solve() {
    int n;
    string s;
    cin >> n >> s;
    if(s[0] == s[n - 1]) cout << "No" << endl;
    else cout << "Yes" << endl;
}
int main() {
    int T;
    cin >> T;
    while(T --) solve();
}

B.#

思维题，答案为从小到大排序后的中间的数。

#include<bits/stdc++.h>
using namespace std;

const int N = 1e6 + 7;
int a[N];
bool cmp(int x, int y){ return x > y;}
void solve() {
    int n;
    cin >> n;
    for(int i = 1; i <= n; i ++) {
        cin >> a[i];
    }
    sort(a + 1, a + n + 1, cmp);
    cout << a[(n + 1) / 2] << endl;
}
int main() {
    int T;
    cin >> T;
    while(T --) solve();
}

D1.#

设 $k$ 为 $max(mex)$，若 $k>m$，则答案为 $k(m+1)$。对于 $[0,k-1]$ 的部分，$f(i)=k$，这部分的总答案为 $k\ast k$。对于 $[k,m]$，$f(i)=i$，这部分的总答案为 $\frac{(k+m)(k-m-1)}{2}$，因此总答案为 $k\ast k+\frac{(k+m)(k-m-1)}{2}$。

D2.#