A. 三人三数

比一下就好了

signed main() {
    IOS;
    int ans = 0;
    int n, a, b, c, d, e, f;
    cin >> n >> a >> b >> c >> d >> e >> f;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            for (int k = 1; k <= n; k++) {
#define ch(x, y) min(abs(x - y), n - abs(x - y))
                if (ch(a, i) <= 2 && ch(b, j) <= 2 && ch(c, k) <= 2) {
                    ans++;
                    continue;
                }
                if (ch(d, i) <= 2 && ch(e, j) <= 2 && ch(f, k) <= 2)
                {
                    ans++;
                }
            }
        }
    }
    cout << ans << endl;
    return 0;
}

B. 最后派对

模拟写一下就好了

struct node {
    int v, w;
};
bool cmp(node a, node b) {
    if ((a.v + a.w) == (b.v + b.w)) {
        return a.v < b.v;
    }
    return (a.v + a.w) < (b.v + b.w);
}
signed main()
{
    IOS;
    int n, m;
    cin >> n >> m;
    vector<node> t(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> t[i].v >> t[i].w;
    }
    sort(t.begin() + 1, t.end(), cmp);
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        int tmp = t[i].w + t[i].v;
        if (sum + tmp <= m) {
            sum += tmp;
        }
        else if(sum + t[i].v / 2  + t[i].w <= m) {
            cout << i << endl;
            return 0;
        }
        else {
            cout << i - 1 << endl;
            return 0;
        }
    }
    cout << n << endl;
    return 0;
}

C. 边走边种

离散化，找好边界

int n;
#define MAX 1000005
int L[MAX], R[MAX];
int idx;
vector<int> v;

signed main()
{
    IOS;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int x;
        char c;
        cin >> x >> c;
        int tmp = c == 'R' ? idx + x : idx - x;
        L[i] = min(tmp, idx) + 1, R[i] = max(tmp, idx);
        v.push_back(L[i]), v.push_back(R[i] + 1);
        idx = tmp;
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    vector<int> len(v.size() + 1);
    for (int i = 1; i <= n; i++)
    {
        L[i] = lower_bound(v.begin(), v.end(), L[i]) - v.begin();
        R[i] = lower_bound(v.begin(), v.end(), R[i] + 1) - v.begin();
        len[L[i]]++, len[R[i]]--;
    }
    int ans = 0;
    for (int i = 0; i < v.size(); i++)
    {
        if (i) len[i] += len[i - 1];
        if (len[i] >= 2)
        {
            ans += v[i + 1] - v[i];
        }
    }
    cout << ans << endl;

    return 0;
}

D. 可爱序列

取模做好DP就好了

#define MAX 1000005
// vector<int> q[MAX];
int idx[MAX];
int DP[MAX];

signed main()
{
    IOS;
    // cout << 5 % 1 << endl;
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        x %= m;
        int tmp = m - x;
        tmp %= m;
        DP[i] = idx[tmp] + 1;
        idx[x] = max(idx[x], DP[i]);
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        // cout << DP[i] << endl;
        ans = max(ans, DP[i]);
    }
    cout << n - ans << endl;

    return 0;
}

E. 立方方方

很经典的莫队题，只要能增删一个数并且可以离线操作就可以莫队

#define MAX 1000005
int n, s, q, a, l, r;
vector<int> p;
vector<int> ep[MAX];
struct node
{
    int uid, l, r;
} ans[MAX];
bool cmp(const node A, const node B)
{
    if (A.l / s != B.l / s)
        return A.l / s < B.l / s;
    return ((A.l / s) & 1) ? A.r < B.r : A.r > B.r;//不这么写会T
}
int now = 0, cnt[MAX], res[MAX];
bool isprime[MAX];
void add(int x)
{
    for (int &i : ep[x])
    {
        int t = ++cnt[i] % 3;
        if (t == 1)
            now++;
        if (t == 0)
            now--;
    }
}
void del(int x)
{
    for (int &i : ep[x])
    {
        int t = --cnt[i] % 3;
        if (t == 2)
            now++;
        if (t == 0)
            now--;
    }
}

signed main()
{
    IOS;
    for (int i = 2; i <= 1000; i++)
    {
        if (!isprime[i])
        {
            for (int j = i * 2; j <= 1000; j += i)
            {
                isprime[j] = 1;
            }
            p.push_back(i);
        }
    }
    cin >> n >> q;
    s = max(1ll, (ll)sqrt((ll)n * n / q));
    for (int i = 1; i <= n; i++)
    {
        cin >> a;
        for (int j : p)
        {
            while (a % j == 0)
            {
                ep[i].push_back(j);
                a /= j;
            }
            if (a == 1)
                break;
        }
        if (a > 1)
            ep[i].push_back(a);
    }
    for (int i = 1; i <= q; i++)
    {
        cin >> ans[i].l >> ans[i].r;
        ans[i].uid = i;
    }
    sort(ans + 1, ans + 1 + q, cmp);
    int l = 0, r = -1;
    for (int i = 1; i <= q; i++)
    {
        while (r < ans[i].r)
            add(++r);
        while (l > ans[i].l)
            add(--l);
        while (r > ans[i].r)
            del(r--);
        while (l < ans[i].l)
            del(l++);
        res[ans[i].uid] = now;
    }
    for (int i = 1; i <= q; i++)
    {
        if (res[i])
            cout << "No" << endl;
        else
            cout << "Yes" << endl;
    }

    return 0;
}