A. 调饮料师

直接暴力就好了

signed main() {
    IOS;
    int c[3], s[3];
    for (int i = 0; i < 3; i++) {
        cin >> c[i] >> s[i];
    }
    for (int i = 0; i < 100; i++) {
        int idx = i % 3;
        int tmp = min(c[(idx + 1) % 3] - s[(idx + 1) % 3], s[idx]);
        s[idx] -= tmp;
        s[(idx + 1) % 3] += tmp;
    }
    for (int i = 0; i < 3; i++) {
        cout << s[i] << endl;
    }
        return 0;
}

B. 心直口筷

差分一下就好了

#define MAX 2000
int a[MAX];
int n;
signed main() {
    IOS;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        int l, r, s;
        cin >> l >> r >> s;
        a[l] += s, a[r + 1] -= s;
    }
    int res = a[0];
    for (int i = 1; i <= 1999; i++) {
        a[i] += a[i - 1];
        res = max(res, a[i]);
    }
    cout << res << endl;

    return 0;
}

C. 转送苹果

只需要考虑送来的和回来的相不相同就好了

#define MAX 15
int a[MAX], b[MAX];
set<int> st;
signed main() {
    IOS;
    for (int i = 1; i <= 10; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= 10; i++) {
        cin >> b[i];
    }
    for (int i = 1; i <= 10; i++) {
        for (int j = 1; j <= 10; j++) {
            for (int k = 1; k <= 10; k++) {
                for (int l = 1; l <= 10; l++) {
                    if (i == k || j == l) continue;
                    st.insert(1000 - a[i] + b[j] - a[k] + b[l]);
                }
            }
            st.insert(1000 - a[i] + b[j]);
        }
    }
    st.insert(1000);
    cout << st.size() << endl;

    return 0;
}

D. 蜗人赛跑

用单调队列去优化

#define MAX 1000005
int a[MAX];
int x[MAX], s[MAX];
int L, N, rf, rb;
deque<int> q;
signed main()
{
    IOS;
    cin >> L >> N >> rf >> rb;
    int mx = 0, idx = 0;
    for (int i = 1; i <= N; i++)
    {
        cin >> x[i] >> s[i];
    }
    for (int i = 1; i <= N; i++)
    {
        while (!q.empty() && s[q.back()] < s[i])
        {
            q.pop_back();
        }
        q.push_back(i);
    }
    vector<int> qq;
    qq.push_back(0);
    while (!q.empty())
    {
        qq.push_back(q.front());
        q.pop_front();
    }
    int ans = 0;
    for (int i = 1; i < qq.size(); i++)
    {
        ans += s[qq[i]] * (x[qq[i]] - x[qq[i - 1]]);
    }
    cout << (rf - rb) * ans << endl;

    return 0;
}

E. 数位清除

发现每个状态不超过5000个，一共长度100，DP[i][j]表示到i位第j大及以前的数合法的数量

#define MAX 105
int n, a, len[MAX], pos[MAX], st[MAX], DP[MAX][5005], ans[MAX][5005];
const int mod = 1e9 + 7;
signed main()
{
    IOS;
    cin >> a >> n;
    for (int i = 1; i <= n + 1; i++) {
        int num = a + i - 1;
        while (num) {
            pos[len[i]++] = num % 10;
            num /= 10;
        }
        int sta = 1 << len[i];
        for (int j = 1; j < sta; j++) {
            st[i]++;
            for (int k = len[i] - 1; k >= 0; k--)
            {
                if (!(j & (1 << k)))
                    continue;
                ans[i][st[i]] = ans[i][st[i]] * 10 + pos[k];
            }
        }
        sort(ans[i] + 1, ans[i] + st[i] + 1);
        for (int j = 1; j <= st[i]; j++) {
            int pos = upper_bound(ans[i - 1], ans[i - 1] + st[i - 1] + 1, ans[i][j]) - ans[i - 1] - 1;
            DP[i][j] = (DP[i][j - 1] + DP[i - 1][pos]) % mod;
            if (i == 1)
                DP[i][j] = j;
        }
    }
    cout << DP[n + 1][st[n + 1]] << endl;
    return 0;
}