Codeforces 比赛笔记（1）

1、Codeforces Round #775 (Div. 2, based on Moscow Open Olympiad in Informatics)

A

注意最多只能跳一次。

const int N = 110;
int t, n, l, r, a[N];

int main()
{
    read(t);
    while (t--)
    {
        read(n); l = 1, r = n;
        for (int i = 1; i <= n; ++i) read(a[i]);
        while (l < n && a[l + 1]) ++l;
        while (r > 1 && a[r - 1]) --r;
        print(l > r ? 0 : r - l), pc('\n');
    }
    fwrite(pbuf, 1, pp - pbuf, stdout); return 0;
}

B

考虑踢球次数最多的人没踢完的话剩下的球都得自己踢，踢完了剩下的人传着踢就行。

所以答案就是最大的 $a[i]$ 减去其他 $a[i]$ 后的值跟 $1$ 取 $\max$，注意特判全为 $0$ 的情况。

const int N = 1e5 + 10;
int t, n, a[N];

int main()
{
    read(t);
    while (t--)
    {
        read(n);
        for (int i = 1; i <= n; ++i) read(a[i]);
        nth_element(a + 1, a + 1, a + n + 1, greater<int>());
        if (!a[1]) { pc('0'), pc('\n'); continue; }
        for (int i = 2; i <= n; ++i)
        {
            a[1] -= a[i];
            if (a[1] < 1) break;
        }
        if (a[1] < 1) pc('1'), pc('\n');
        else print(a[1]), pc('\n');
    }
    fwrite(pbuf, 1, pp - pbuf, stdout); return 0;
}

C

由于是曼哈顿距离，直接将答案拆成行和列的分别求解即可。

const int N = 1e5 + 10;
int n, m;
ll ans, cnt[N], sum[N];
vector<int> a[320];

int main()
{
    read(n, m);
    for (int x, i = 1; i <= n; ++i)
    {
        a[i].emplace_back(0);
        for (int j = 1; j <= m; ++j)
            read(x), a[i].emplace_back(x);
    }
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= m; ++j)
            ans += cnt[a[i][j]] * i - sum[a[i][j]];
        for (int j = 1; j <= m; ++j)
            ++cnt[a[i][j]], sum[a[i][j]] += i;
    }
    memset(cnt, 0, sizeof(cnt));
    memset(sum, 0, sizeof(sum));
    for (int j = 1; j <= m; ++j)
    {
        for (int i = 1; i <= n; ++i)
            ans += cnt[a[i][j]] * j - sum[a[i][j]];
        for (int i = 1; i <= n; ++i)
            ++cnt[a[i][j]], sum[a[i][j]] += j;
    }
    print(ans);
    fwrite(pbuf, 1, pp - pbuf, stdout); return 0;
}

D

直接暴力枚举 $y$ 和 $\lfloor\dfrac{x}{y}\rfloor$，判断数组中是否存在 $x$ 但不存在 $\lfloor\dfrac{x}{y}\rfloor$ 即可。

由于枚举的是倍数关系，$\dfrac{n}{1}+\dfrac{n}{2}+\dfrac{n}{3}+\cdots+\dfrac{n}{n}=O(n\log{n})$，所以时间复杂度是正确的。

const int N = 1e6 + 10;
int t, n, c, a[N], cnt[N];

inline bool check(int y)
{
    for (int k = 1; k * y <= c; ++k)
    {
        int l = k * y - 1, r = Min(k * y + y - 1, c);
        if (cnt[r] - cnt[l] && !(cnt[k] - cnt[k - 1])) return false;
    }
    return true;
}

int main()
{
    read(t);
    while (t--)
    {
        read(n, c); bool flag = true;
        for (int i = 1; i <= n; ++i) read(a[i]), ++cnt[a[i]];
        for (int i = 1; i <= c; ++i) cnt[i] += cnt[i - 1];
        sort(a + 1, a + n + 1);
        for (int i = 1; i <= n; ++i)
            if (a[i] != a[i - 1] && !check(a[i])) { flag = false; break; }
        puts(flag ? "Yes" : "No");
        for (int i = 1; i <= c; ++i) cnt[i] = 0;
    }
    fwrite(pbuf, 1, pp - pbuf, stdout); return 0;
}

E

F

2、Codeforces Round #776 (Div. 3)

A

判奇偶性即可。

int main()
{
    int t; read(t);
    while (t--)
    {
        int n; char s[50], c; bool flag = false;
        read(s + 1), n = strlen(s + 1), read(c);
        for (int i = 1; i <= n; ++i)
            if (s[i] == c && !(i - 1 & 1) && !(n - i & 1))
                { flag = true; break; }
        puts(flag ? "YES" : "NO");
    }
    flush_pbuf(); return 0;
}

B

讨论 $\lfloor\dfrac{x}{a}\rfloor$ 与 $x\bmod a$ 最大即可。

inline int f(int x, int a) { return x / a + x % a; }

int main()
{
    int t; read(t);
    while (t--)
    {
        int l, r, a; read(l, r, a);
        print(Max(f(r, a), f(Max(l, r - r % a - 1), a))), pc('\n');
    }
    flush_pbuf(); return 0;
}

C

答案只跟点权之和有关，直接取最小点权即可。

const int N = 2e5 + 10;
struct node { int x, w, id; } a[N];

inline bool cmp1(const node &x, const node &y) { return x.w < y.w; }
inline bool cmp2(const node &x, const node &y) { return x.x < y.x; }

int main()
{
    int t; read(t);
    while (t--)
    {
        int n, m, ans = 0; read(n, m);
        for (int i = 1; i <= m; ++i) read(a[i].x, a[i].w), a[i].id = i;
        sort(a + 1, a + m + 1, cmp1);
        for (int i = 1; i <= n << 1; ++i) ans += a[i].w;
        print(ans, '\n'); sort(a + 1, a + (n << 1) + 1, cmp2);
        for (int i = 1; i <= n; ++i)
            print(a[i].id, ' ', a[(n << 1) - i + 1].id, '\n');
    }
    flush_pbuf(); return 0;
}

D

直接用递归倒推还原即可。

const int N = 4e3 + 10;
int a[N], b[N], d[N];

void solve(int n)
{
    if (n == 1) return;
    for (int i = 1; i <= n; ++i) b[i] = b[n + i] = a[i];
    int pos;
    for (int i = n; i < n << 1; ++i)
        if (b[i] == n) { pos = i; break; }
    d[n] = pos - n;
    for (int i = 1; i < n; ++i) a[i] = b[pos - n + i];
    solve(n - 1);
}

int main()
{
    int t; read(t);
    while (t--)
    {
        int n; read(n);
        for (int i = 1; i <= n; ++i) read(a[i]);
        solve(n);
        for (int i = 1; i <= n; ++i) print(d[i]), pc(' '); pc('\n');
    }
    flush_pbuf(); return 0;
}

E

用一个支持插入删除查询最小最大值的数据结构，注意特判删最后一个点的情况。

const int N = 2e5 + 10;
int a[N];
multiset<int> s;

int main()
{
    int t; read(t);
    while (t--)
    {
        int n, d, ans; read(n, d); ans = 0;
        for (int i = 1; i <= n; ++i) read(a[i]);
        for (int i = 1; i <= n; ++i) s.insert(a[i] - a[i - 1] - 1);
        for (int i = 1; i < n; ++i)
        {
            auto it = s.lower_bound(a[i] - a[i - 1] - 1); s.erase(it);
            it = s.lower_bound(a[i + 1] - a[i] - 1), s.erase(it);
            s.insert(a[i + 1] - a[i - 1] - 1);
            ans = Max(ans, Min(*s.begin(), Max(*--s.end() - 1 >> 1, d - a[n] - 1)));
            it = s.lower_bound(a[i + 1] - a[i - 1] - 1), s.erase(it);
            s.insert(a[i] - a[i - 1] - 1), s.insert(a[i + 1] - a[i] - 1);
        }
        auto it = s.lower_bound(a[n] - a[n - 1] - 1); s.erase(it);
        ans = Max(ans, Min(*s.begin(), Max(*--s.end() - 1 >> 1, d - a[n - 1] - 1)));
        print(ans), pc('\n'); multiset<int>().swap(s);
    }
    flush_pbuf(); return 0;
}

F

G

3、Educational Codeforces Round 124 (Rated for Div. 2)

4、Codeforces Round #777 (Div. 2)

To be continued...