AtCoder Beginner Contest 240

A

#include <bits/stdc++.h>
using namespace std;
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b)  swap(a, b);
    if (b - a == 1 || b - a == 9)   puts("Yes");
    else puts("No");
    return 0;
}

B

#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
    scanf("%d", &n);
    set<int> s;
    for (int i = 1; i <= n; i ++ ) {
        int x;
        scanf("%d", &x);
        s.insert(x);
    }
    printf("%d\n", s.size());
    return 0;
}

C

#include <bits/stdc++.h>
using namespace std;
int n, m;
int st[10010];
int main() {
    scanf("%d%d", &n, &m);
    st[0] = 1;
    for (int i = 1; i <= n; i ++ ) {
        int a, b;
        scanf("%d%d", &a, &b);
        vector<int> c(m + 10);
        for (int j = 0; j <= m; j ++ ) 
            if (st[j] == 1)  {
                if (j + a <= m) c[j + a] = 2;
                if (j + b <= m) c[j + b] = 2;
            }
        for (int j = 0; j <= m; j ++ )
            if (c[j] == 2) st[j] = 1;
            else st[j] = 0;
    }
    if (st[m] == 1) puts("Yes");
    else puts("No");
    return 0;
}

D

#include <bits/stdc++.h>
#define PII pair<int, int>
using namespace std;
int n, len;
int main() {
    scanf("%d", &n);
    vector<PII> v;
    for (int i = 1; i <= n; i ++ )  {
        int x;
        scanf("%d", &x);
        len ++;
        if (v.size() == 0)  v.push_back({x, 1});
        else {
            if (v[v.size() - 1].first == x && v[v.size() - 1].second == x - 1) 
                len -= x, v.pop_back();
            else {
                if (v[v.size() - 1].first == x) v[v.size() - 1].second ++;
                else v.push_back({x, 1});
            }
        }
        printf("%d\n", len);
    }
    return 0;
}

E

#include <bits/stdc++.h>
#define PII pair<int, int>
using namespace std;
const int N = 2e5 + 10, M = N * 2;
int n;
PII range[N];
int h[N], e[M], ne[M], idx, s[N];
void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
void dfs(int u, int fa)
{
    s[u] = 0;
    int sum = 1;
    for (int i = h[u]; i != -1; i = ne[i]) {
        int j = e[i];
        if (j == fa)    continue;
        dfs(j, u);
        s[u] += max(s[j], 1);
    }
}
void dfs1(int u, int fa, int l, int r)
{
    int ll = l, rr = r;
    range[u].first = l, range[u].second = r;
    for (int i = h[u]; i != -1; i = ne[i]) {
        int j = e[i];
        if (j == fa)    continue;
        dfs1(j, u, ll, ll + max(s[j], 1) - 1);
        ll += max(s[j], 1);
    }
}
int main() {
    scanf("%d", &n);
    memset(h, -1, sizeof h);
    for (int i = 1; i < n; i ++ ) {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b), add(b, a);
    }
    dfs(1, -1);
    dfs1(1, -1, 1, s[1]);
    for (int i = 1; i <= n; i ++ ) 
        printf("%d %d\n", range[i].first, range[i].second);
    return 0;
}

F
我们可以发现，每次连续的\(y_i\)个数的值是相等的，所以我们可以一段一段的计算

\[首先定义 \quad z_i = \sum_{k=1}^{i}y_k \quad 设此段为连续的个数为n个x_i那么我们可以得到当前一段的和为 n \times x_i \\ 所以可以得到 B_{z_{i-1}+n} = B_{z_{i-1}} + n \times x_i \\ 同理可以得到A_{z_{i-1}+n} = A_{z_{i-1}} + \sum_{k=1}^{n} B_{z_{i-1}+k} \\ A_{z_{i-1}+n} = A_{z_{i-1}} + \sum_{k=1}^{n}(B_{z_{i-1}} + k \times x_i) \\ A_{z_{i-1}+n} = A_{z_{i-1}} + B_{z_{i-1}} \times n + x_i \times \frac{n \times (n + 1)}{2} \]

至此可以发现每段的值都可以由前面一段得到,而观察上式发现是一个关于n的一元二次方程

\[A_{z_{i-1}+n} = A_{z_{i-1}} + B_{z_{i-1}} \times n + x_i \times \frac{n \times (n + 1)}{2} \\ 如果前一段的B_{z_{i-1}} > 0时，而x_i<0时，发现上式是一个关于n的一元二次方程当n = \frac{b}{-2a}取得最大值 \]

#include <bits/stdc++.h>
#define LL long long
using namespace std;
int T, n, m;
void solve()
{   
    scanf("%d%d", &n, &m);
    LL a = 0, b = 0, ans = -1e18;  //a表示A的前缀和，b表示B的前缀和
    for (int i = 1; i <= n; i ++ ) {
        LL x, y;
        scanf("%lld%lld", &x, &y);
        ans = max(ans, a + b + x);  //当a = 0, b = 0, 所有数都为负数时，可以取得最大值为第一个数
        if (b > 0 && b + x * y < 0) {
            LL k = b / -x;
            ans = max(ans, a + b * k + k * (k + 1) / 2 * x); //最大值在对称轴取得
        }
        a += b * y + (y + 1) * y / 2 * x;
        b += x * y;
        ans = max(ans, a);
    }
    printf("%lld\n", ans);
}
int main() {
    scanf("%d", &T);
    while (T -- )   solve();
    return 0;
}