A
双指针

#include <bits/stdc++.h>
#define PII pair<int, int>
using namespace std;
const int N = 2e5 + 10;
int T, n, d;
int a[N];
int main() {
    scanf("%d", &T);
    while (T -- ) {
        scanf("%d", &n);
        d = -1;
        vector<PII> ans;
        for (int i = 1; i <= n; i ++ )  cin >> a[i];
        for (int i = 1; i <= n; ) {
            int j = i + 1;
            while (j <= n && a[j] >= a[j - 1])  j ++;
            d = max(d, a[j - 1] - a[i]);
            i = j;
        }
        for (int i = 1; i <= n; ) {
            int j = i + 1;
            while (j <= n && a[j] >= a[j - 1])  j ++;
            if (d == a[j - 1] - a[i])   ans.push_back({i, j - 1});  
            i = j;
        }
        for (auto x : ans)
            printf("%d %d ", x.first, x.second);
        puts("");
    }
    return 0;
}

B
暴力

#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int T, n, m;
char s[N];
char a[N][N];
int main() {
    scanf("%d", &T);
    while (T -- ) {
        int ans = 0;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i ++ ) {
            scanf("%s", s + 1);
            for (int j = 1; j <= m; j ++ ) 
                a[i][j] = s[j];
        }
        for (int i = 1; i <= n; i ++ ) 
            for (int j = 1; j <= m; j ++ ) 
                if (a[i][j] == '*') 
                    for (int i1 = -1; i1 <= 1; i1 ++ ) 
                        for (int j1 = -1; j1 <= 1; j1 ++ ) 
                            if (i + i1 >= 1 && i + i1 <= n && j + j1 <= m && j + j1 >= 1 && a[i + i1][j + j1] != '*')
                                a[i + i1][j + j1] = 'Z';
        for (int i = 1; i <= n; i ++ ) 
            for (int j = 1; j <= m; j ++ ) 
                if (a[i][j] == 'P')
                    ans ++;
        printf("%d\n", ans);
    }
    return 0;
}

C
根据题意模拟

#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int T, n, r, g, ex;
int m, m1, m2;
int main() {
    scanf("%d", &T);
    while (T -- ) {
        scanf("%d%d.%d", &n, &m1, &m2);
        ex = 0;
        m = (m1 - 1) * 100 + m2 * 10;
        while (n) {
            r = n * 100, g = min(10000, n * m);
            n = r / 200, ex += r / 10;
            ex += g / 10;
        } 
        printf("%d\n", ex);
    }
    return 0;
}

D
简单数学

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MOD = 998244353;
int T, p, ans;
int len1, len2;
string s1, s2;
signed main() {
    cin >> T;
    while (T -- ) {
        p = 1;
        ans = 0;
        cin >> s1 >> s2;
        len1 = s1.size(), len2 = s2.size();
        reverse(s1.begin(), s1.end());
        reverse(s2.begin(), s2.end());
        for (int i = 0; i < len1; i ++ ) {
            int k = s1[i] - '0';
            ans = (ans + k * p % MOD * len2) % MOD;
            p = (p * 10) % MOD;
        }
        p = 1;
        for (int i = 0; i < len2; i ++ ) {
            int k = s2[i] - '0';
            ans = (ans + k * p % MOD * len1) % MOD;
            p = (p * 10) % MOD;
        }
        cout << ans << endl;
    }
    return 0;
}

E
找到所有黑点的数量后，\(C_{cnt}^{2}\)

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 3e6 + 10;
int T, n, res;
bool st[N];
int h[N], e[N * 2], ne[N * 2], idx;
struct node {
    int ver, d;
};
signed main() {
    scanf("%lld", &T);
    while (T -- ) {
        scanf("%lld", &n);
        int max_d = 1;
        idx = 0, res = 0;
        for (int i = 1; i <= n; i ++ )  st[i] = 0;
        for (int i = 1; i <= n; i ++ )  h[i] = -1;
        function<void(int, int)> add = [&](int a, int b) {
            e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
        };
        for (int i = 1; i < n; i ++ ) {
            int a, b;
            scanf("%lld%lld", &a, &b);
            add(a, b), add(b, a);
        }
        function<void()> bfs = [&]() {
            queue<node> q;
            q.push({1, 1});
            st[1] = 1;
            while (q.size()) {
                auto t = q.front();
                q.pop();
                int u = t.ver, step = t.d;
                if (step % 2 == 1)    res ++;
                for (int i = h[u]; i != -1; i = ne[i]) {
                    int j = e[i];
                    if (st[j])    continue;
                    q.push({j, step + 1});
                    st[j] = true;
                }
            }
        };
        bfs();
        printf("%lld\n", res + res * (res - 1) / 2);
    }
    return 0;
}

F
贪心，如果最长的数量小于总长的一半则所有点都成立
否则，只有在最长链上的点才成立，用最长链的长度减去两端不满足条件的即可

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e5 + 10;
int T, n, sum, maxx;
int a[N];
signed main() {
    scanf("%lld", &T);
    while (T -- ) {
        scanf("%lld", &n);
        sum = 0, maxx = 0;
        for (int i = 1; i <= n; i ++ )  
            scanf("%lld", &a[i]), sum += a[i], maxx = max(maxx, a[i]);
        if (maxx <= sum - maxx)
            printf("%lld\n", sum);
        else
            printf("%lld\n", maxx - (maxx - sum / 2 - 1) * 2);
    }
    return 0;
}