首先是A题，看错题目了，本来应该计算大于50，结果算了小于等于50的。

下面是正解代码：

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

int mon2day(int y, int m) {
    if (m & 1 && m < 8)return 31;
    if (!(m & 1) && m >= 8)return 31;
    if (m == 2)
        return (y % 4 == 0) ? 29 : 28;
    else
        return 30;
}

int ch2num[10] = {13, 1, 2, 3, 5, 4, 4, 2, 2, 2};

struct Date {
    int year, mon, day;

    auto operator++() {
        day++;
        if (day > mon2day(year, mon))
            day = 1, mon++;
        if (mon > 12)
            mon = 1, year++;
        return *this;
    }

    int flatten() const {
        return year * 10000 + mon * 100 + day;
    }

    int getSum() const {
        int date = flatten();
        int sum = 0;
        while (date) {
            sum += ch2num[date % 10];
//          cout << date % 10 << " " << sum << endl;
            date /= 10;
        }
        return sum;
    }

    auto operator<=>(const Date&b) const {
        return flatten() <=> b.flatten();
    }
};

int main() {
    int ans = 0;
    for (auto d = Date{2000, 1, 1}; d <= Date{2024, 4, 13}; ++d) {
        int sum = d.getSum();
        cout << d.flatten() << " " << sum << endl;
        if (sum > 50) ans++;
    }
    cout << ans << endl;

    return 0;
}

接下来C题，二分没什么问题。（由于比较菜，写不好二分，所以直接用STL了）

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

constexpr int N = 1e6 + 5;
int n, S;
int p[N], c[N];

// cost的缓存
int cost[N];

// 与第gnum次组团训练相当的代价
int getCost(int gnum) {
    if (cost[gnum] != 0) return cost[gnum];
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        if (c[i] >= gnum )sum += p[i];
    }
    return cost[gnum] = sum;
}

// 进行ss次组团训练的情况，用来排序
// 存的数据是ss，排序指标是getCost(ss)
struct Case {
    int ss;
    auto operator<(auto b) const {
        return getCost(ss) > getCost(b.ss);
    }
    auto operator==(auto b) const {
        return getCost(ss) == getCost(b.ss);
    }
};

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cin >> n >> S;

    int max_c = 0;
    for (int i = 1; i <= n; i++) {
        cin >> p[i] >> c[i];
        max_c = max(max_c, c[i]);
    }

    vector<Case> vec;
    for (int i = 1; i <= max_c; i++) {
        vec.push_back({i});
        // cout << "S_num=" << i << ", S_cost=" << getCost(i) << endl;
    }

    auto target = Case{0};
    cost[0] = S;

    auto l = lower_bound(vec.begin(), vec.end(), target);
    int group_num = max_c;
    if (l != vec.end())
        group_num = l->ss - 1;

    int ans = S * group_num;
    for (int i = 1; i <= n; i++) {
        c[i] = max(0LL, c[i] - group_num);
        ans += c[i] * p[i];
    }

    cout << "ans = " << ans << endl;

    return 0;
}