第 45 届国际大学生程序设计竞赛（ICPC）亚洲区域赛（沈阳）补题记录

比赛链接：Here

很可惜，如果再强一点，就可以拿牌子了。

5道即可金牌尾 or 银首

F. Kobolds and Catacombs (思维)

真不难，只是理解错了题意

如果原数组 (a) 和 排序后的数组 (b) 在某个位置前缀和相同和可以划分为一组

const int N = 1e6 + 10;
ll a[N], b[N];
int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int n; cin >> n;
    for (int i = 1 ; i <= n; ++i) cin >> a[i], b[i] = a[i];
    sort(a + 1, a + 1 + n);
    ll sa = 0, sb = 0, cnt = 0;
    for (int i = 1; i <= n; ++i) {
        sa += a[i], sb += b[i];
        if (sa == sb) cnt ++;
    }
    cout << cnt;
}

G. The Witchwood (签到)

递减排序然后累加前 (k) 个即可

H. The Boomsday Project

淦，很考验细节处理的单调队列优化DP，当初没学过，不知道怎么处理

const int N = 5e2 + 10;
int d[N], k[N], c[N];
struct Node {int p, q;} node[100005];
bool cmp(Node a, Node b) {return a.p < b.p;}
int t[100005 * 3]; //第几次是在第几天
ll dp[100005 * 3];

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int n, m, r;
    cin >> n >> m >> r;
    for (int i = 1; i <= n; ++i) cin >> d[i] >> k[i] >> c[i];
    int w = 0;
    for (int i = 1; i <= m; ++i) cin >> node[i].p >> node[i].q;
    sort(node + 1, node + 1 + m, cmp);
    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= node[i].q; ++j) t[w + j] = node[i].p;
        w += node[i].q;
    }
    t[0] = t[1];
    dp[0] = 0;
    for (int i = 1; i <= w; ++i) dp[i] = dp[i - 1] + r;
    deque<int>q[n + 1];
    for (int i = 1; i <= n; ++i) q[i].push_back(0);
    for (int i = 1; i <= w; ++i) {
        for (int j = 1; j <= n; ++j) {
            while (!q[j].empty() and (t[i] - t[q[j].front() + 1] >= d[j] || i - q[j].front() > k[j]))
                q[j].pop_front();
            dp[i] = min(dp[i], dp[q[j].front()] + c[j]);
        }
        for (int j = 1; j <= n; ++j) {
            while (!q[j].empty() and dp[i] <= dp[q[j].back()]) q[j].pop_back();
            q[j].push_back(i);
        }
    }
    cout << dp[w] << '
';
}

I. Rise of Shadows

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    ll h, m, a;
    cin >> h >> m >> a;
    ll d = __gcd(h - 1, m);
    cout << min(d * (2 * (a / d) + 1), h * m);
}

K . Scholomance Academy

真 · 阅读理解题

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int n; cin >> n;
    int s1 = 0, s2 = 0;
    vector<int>v1, v2;
    for (int i = 1; i <= n; ++i) {
        char c; int x;
        cin >> c >> x;
        if (c == '+') s1++, v1.push_back(x);
        else s2++, v2.push_back(x);
    }
    sort(v1.begin(), v1.end());
    sort(v2.begin(), v2.end());
    double ans = 0;
    for (int i = 0; i < s1; ++i) {
        ans += lower_bound(v2.begin(), v2.end(), v1[i]) - v2.begin();
    }
    ans = ans / s1 / s2;
    cout << fixed << setprecision(10) << ans;
}