两日总结十

比赛总结：

Educational Codeforces Round 133 (Rated for Div. 2)

D题实在是有些可惜，当时都已经想到了，就是时间卡了一点点，不用队列就直接过了，直接求出K的最大值632，然后就是直接暴力就好了。。。一开始想的有点复杂。

#include 
#define OST ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define endl "\n"
#define ll long long
#define pair pair
using namespace std;
const int mod = 998244353;
int vis[200005];
void solve() {
    int n, k;
    cin >> n >> k;
    vector f(n + 1, 0), ff(n + 1, 0), fff(n + 1, 0);
    queue q;
    for (int i = k; i <= n; i += k) {
        q.push({i, k + 1});
        f[i] = 1;
    }
    while (!q.empty()) {
        auto x = q.front();
        q.pop();
        if (x.second >= 633) break;
        if (x.second > k) {
            fff = ff;
            ff  = f;
            k   = x.second;
            for (int i = 1; i <= n; i++) vis[i] = 0;
        }
        for (int i = x.first + x.second; i <= n; i += x.second) {
            f[i] += (ff[x.first] - fff[x.first] + mod) % mod;
            f[i] %= mod;
            if (!vis[i]) q.push({i, x.second + 1}), vis[i] = 1;
        }
    }
    for (int i = 1; i <= n; i++) { cout << f[i] << " "; }
    cout << endl;
}
 
signed main() {
    int T = 1;
    // cin >> T;
    OST;
    while (T--) { solve(); }
    return 0;
}

 后面改的代码AC:

#include 
#define OST ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define endl "\n"
#define ll long long
#define pair pair
using namespace std;
const int mod = 998244353;
void solve() {
    int n, k;
    cin >> n >> k;
    vector f(n + 1, 0), ff(n + 1, 0), ans(n + 1, 0);
 
    f[0]   = 1;
    ll sum = 0;
    for (int step = 0; step < 633; step++) {
        sum += step;
        if (sum > n) break;
        for (int j = 0; j + k + step <= n; j++) { ff[j + k + step] = f[j] + ff[j], ff[j] %= mod; }
        for (int j = 0; j <= n; j++) f[j] = ff[j], ff[j] = 0, ans[j] += f[j], ans[j] %= mod;
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << endl;
    cout << endl;
}
 
signed main() {
    int T = 1;
    // cin >> T;
    OST;
    while (T--) { solve(); }
    return 0;
}

真的难受！！！

然后就是那个C题：想到很简单主要是划分值比较复杂，和容易错，而且在那个位置划分也要考虑，是真的难写。。。。

#include 
#define OST ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define endl "\n"
using namespace std;
#define ll long long
const int N  = 2e5 + 5;
const ll inf = 0x3f3f3f3f3f3f3f3f;
ll t[N][2], f[N][2], houjia[N][2], houjian[N][2];
void solve() {
    int n;
    cin >> n;
    for (int i = 0; i <= 1; i++)
        for (int j = 0; j < n; j++) cin >> t[j][i], t[j][i]++;
    t[0][0]--;
 
    for (int i = 0; i < 2; i++) {
        houjia[n][i]  = -inf;
        houjian[n][i] = -inf;
        for (int j = n - 1; j >= 0; j--) {
            houjia[j][i]  = max(houjia[j + 1][i] + 1, t[j][i]);
            houjian[j][i] = max(houjian[j + 1][i] - 1, t[j][i]);
        }
    }
    ll step = 0, res = inf, ans = inf, ma = 0, flag = 0;
    for (int i = 0; i < n; i++) {
        ma  = max(t[i][flag] - step, ma);
        res = max({ma, houjian[i][flag] - step, houjia[i][flag ^ 1] - (step + 2 * (n - i - 1) + 1)});
        ans = min(res, ans);
        // cout << houjian[flag][i] - step << "--" << step << endl;
        step++;
        flag ^= 1;
        ma  = max(t[i][flag] - step, ma);
        res = max({ma, houjian[i][flag] - step, houjia[i + 1][flag ^ 1] - (step + 2 * (n - i - 1))});
        ans = min(ans, res);
        step++;
    }
    cout << 2 * n - 1 + max(0ll, ans) << endl;
}
 
 
signed main() {
    int T = 1;
    cin >> T;
    while (T--) { solve(); }
    return 0;
}

航电杯6

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