【21.10.24】The 15th Chinese Northeast Collegiate Programming Contest题解

最新推荐文章于 2024-05-05 15:17:58 发布

原创

最新推荐文章于 2024-05-05 15:17:58 发布 · 724 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#算法

本文档展示了使用C++实现的组合数计算函数，并结合数学原理，探讨了如何利用递归和预计算方法优化计算过程。同时涵盖了线性代数中的Matrix操作，以及动态规划在Vertex Deletion问题中的应用。

A. Matrix（组合数+数学）

#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
ll fac[100010], inv[100010];
const int maxn = 3e7 + 10;
ll mul[maxn];
 
ll kasumi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) {
            ans = ans * a % mod;
        }
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
 
void zuheshu() {
    fac[0] = 1;
    for (int i = 1; i <= 100000; i++) {
        fac[i] = fac[i - 1] * i % mod;
    }
    inv[100000] = kasumi (fac[100000], mod - 2);
    for (int i = 99999; i >= 0; i--) {
        inv[i] = inv[i + 1] * (i + 1) % mod;
    }
}
 
void jiechen() {
    mul[0] = 1;
    mul[1] = 1;
    for (int i = 2; i <= maxn - 10; i++) {
        mul[i] = mul[i - 1] * i % mod;
    }
}
 
ll get_c (ll x, ll y) {
    return fac[x] * inv[y] % mod * inv[x - y] % mod;
}
 
int main() {
    int T;
    scanf ("%d", &T);
    zuheshu();
    jiechen();
    while (T--) {
        ll n;
        scanf ("%lld", &n);
        ll ans = 0;
        ll p = mul[n];
        ll q = mul[n * n - n];
        for (int i = 1; i <= n; i++) {
            ans += n * get_c (n * n - i, n - 1) * p * q;
        }
        cout << ans << endl;
    }
}

B.Cypher（大模拟）

有空填坑

C.Vertex Deletion(简单树形DP)

#include <btis/stdc++.h>
using namespace std;
#define mod 998244353
const int maxn = 2e5 + 50;
 
ll dp[maxn][5];
 
vector<int>G[maxn];
 
void dfs (int u, int father) {
    dp[u][0] = 1;
    dp[u][1] = 1;
    dp[u][2] = 1;
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (v == father) {
            continue;
        } else {
            dfs (v, u);
            dp[u][0] = (dp[u][0] % mod * (dp[v][0] + dp[v][2]) % mod) % mod;
            dp[u][1] = dp[u][1] * dp[v][0] % mod;
            dp[u][2] = (dp[u][2] * (dp[v][0] + dp[v][1] + dp[v][2]) % mod) % mod;
        }
    }
    dp[u][2] = (dp[u][2] - dp[u][1] + mod) % mod;
}
 
int main() {
    int T;
    scanf ("%d", &T);
    while (T--) {
        int n;
        scanf ("%d", &n);
        for (int i = 1; i <= n - 1; i++) {
            int u, v;
            scanf ("%d%d", &u, &v);
            G[u].push_back (v);
            G[v].push_back (u);
        }
        dfs (1, 0);
        cout << (dp[1][0] + dp[1][2]) % mod << endl;
        for (int i = 0; i <= n; i++) {
            G[i].clear();
        }
    }
}

D.Lowbit(线段树区间求和)

注意：

lowbit不能取模，会改变下一轮lowbit的答案

#include <bits/stdc++.h>
#define ll long long
#define lson root << 1
#define rson root << 1 | 1
#define lowbit(x) ((

最低0.47元/天解锁文章