A

200/8=25

B

#include <bits/stdc++.h>
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 1e6 + 7;
bool isPrime(int n) {
    for (int i = 2, x = sqrt(n); i <= x; ++i) {
        if (n % i == 0) return 0;
    }
    return 1;
}
bool ok(int n) {
    while (n) {
        if (n % 10 != 2 and n % 10 != 3 and n % 10 != 5 and n % 10 != 7) {
            return 0;
        }
        n /= 10;
    }
    return 1;
}
int main() {
    int cnt = 0;
    rep(i, 2, 20210605) if (ok(i) and isPrime(i))++ cnt;
    printf("%d\n", cnt);  // 1903
    return 0;
}

C

#include <bits/stdc++.h>
#define rep(i, l, r) for (int i = l; i <= r; ++i)
typedef long long ll;
using namespace std;
const int N = 5e5 + 7;
int nextDay(int n) {
    int y = n / 10000;
    int d = n % 100;
    int m = n / 100 % 100;
    if (m == 12 and d == 31)
        y += 1, m = 1, d = 1;
    else if (m == 1 || m == 3 || m == 5 || m == 7 || m == 8 || m == 10 ||
             m == 12) {
        if (d == 31)
            m += 1, d = 1;
        else
            ++d;
    } else if (m == 2) {
        if (y % 4 == 0) {
            if (d == 29)
                m += 1, d = 1;
            else
                d += 1;
        } else {
            if (d == 28)
                m += 1, d = 1;
            else
                d += 1;
        }
    } else {
        if (d == 30)
            m += 1, d = 1;
        else
            d += 1;
    }
    return y * 10000 + m * 100 + d;
}
bool chk(int n) {
    int res = 0;
    while (n) {
        res += n % 10;
        n /= 10;
    }
    int t = sqrt(n);
    return res == t * t;
}
int main() {
    int cnt = 0;
    for (int i = 20010101; i <= 20211231; i = nextDay(i))
        if (chk(i)) ++cnt;
    cout << cnt << endl;  // 977
    return 0;
}

D

E

#include <bits/stdc++.h>
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
const int N = 1e6 + 7;
int main() {
    string s;
    cin >> s;
    for (char &c : s)
        if (islower(c)) c += 'A' - 'a';
    cout << s;
    return 0;
}

F

G

#include <bits/stdc++.h>
#define rep(i, l, r) for (int i = l; i <= r; ++i)
using namespace std;
typedef long long ll;
ll f[65][55];
int num[65], k;
ll dfs(int len, int cnt, bool limit) {
    if (!len) return cnt == k;  // 到达终态，判断
    if (cnt > k) return 0;      // 剪枝
    if (!limit && ~f[len][cnt])
        return f[len][cnt];  // 如果未达到终态，且已计算过，即可复用
    ll &x = f[len][cnt], ans(0);
    int ed = limit ? num[len] : 1;  // 边界
    for (int i = 0; i <= ed; ++i) {
        ans += dfs(len - 1, cnt + (i == 1), limit && i == ed);
    }
    if (!limit) x = ans;  // 记录复用
    return ans;
}
int main() {
    int len(0);
    ll n;
    scanf("%lld%d", &n, &k);
    memset(f, -1, sizeof f);
    while (n) {
        num[++len] = n & 1;
        n >>= 1;
    }
    cout << dfs(len, 0, true) << '\n';
    return 0;
}