2024 (ICPC) Jiangxi Provincial Contest(VP补题记录)

已ac 8/12，赛时7题，赛后1题。

A(签到中的签到，pass)

C(简单思维)

要么 n 要么 n-1.

FastIO 已省略

def solve():n,s = MI()a = LI()print1(n if sum(a) == s else n-1)# for _ in range(I()):
solve()

G

给你一个长度不超过 $10^{14}$ 的11进制数，判断是不是5的倍数，由于 $11^i\mathrm{mod}10=1$，所以所有位累加即可。

mod = 5
dic = {str(i):i for i in range(10)}
dic['A'] = 10
def solve():n = I()ans = 0for i in range(n):x,y = MS()ans += int(x) * dic[y]ans %= 5print('Yes' if ans % 5 == 0 else 'No')for _ in range(I()):solve()# print(set([pow(11,i,10) for i in range(100000)]))

J(按题意模拟即可)

se = set()
for c in "psm":se.add('1'+c)se.add('9'+c)
for i in range(1,8):se.add(str(i) + 'z')
def solve():s = S()dic = {}for i in range(0,28,2):ts = s[i:i+2]dic[ts] = dic.get(ts,0)+1flag = 1for v in dic.values():if v != 2:flag = 0breakif flag:print1("7 Pairs")returnif se == set(dic.keys()): print1("Thirteen Orphans")else: print1("Otherwise")for _ in range(I()):solve()

K

显然有m-1种分叉情况，所以 $ans=2^{m-1}$ % $mod$ .

mod = 998244353
def solve():m = I()print1(pow(2, m-1, mod))solve()

H(卷积加权和反过来看)

考虑把卷积核 $K$ 加权到输入矩阵 $I$ 的过程反过来看，判断卷积核 $K$ 的每一个元素会扫到哪些 $I$ (即局部链接了哪些ceil). 用个二维前缀和维护一下即可。

def solve():n,m,k,l = MI()ls = [[0]*(m+1)]for i in range(n):ls.append([0]+LI())for i in range(1,n+1):for j in range(1,m+1):ls[i][j] = ls[i][j] + ls[i-1][j] + ls[i][j-1] - ls[i-1][j-1]ans = 0for i in range(1,k+1):for j in range(1,l+1):tmp = ls[n-k+i][m-l+j] - ls[n-k+i][j-1] - ls[i-1][m-l+j] + ls[i-1][j-1]ans += abs(tmp)print1(ans)solve()

L. Campus

假设 $k$ 个出口开放区间最多组成 $len(seq)$ 个不同情况的时间区间(出口不一样)，考虑每个区间跑一次 $dijkstra$ 即可, $k\le 15$。注：py会TLE。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e5+5;int n,m,tot,k,t;
int ver[N+N], head[N], edge[N+N], nxt[N+N], a[N];
ll ans[N], inf = 1e15;
ll dist[N];
int p[20], l[20], r[20], node[20];
bool vis[N];
int ls[N];
vector<pii> seq;void solve(){auto add = [&](int x,int y,int z){ver[++tot]=y,edge[tot]=z;nxt[tot]=head[x],head[x]=tot;};auto dijkstra = [&](int node[]){priority_queue<pair<int, int> > q;for(int i = 1; i <= k; i++){if (node[i]){dist[p[i]] = 0;q.push({-0,p[i]});}}while(!q.empty()){pii tmp = q.top();q.pop();int x = tmp.second;if(vis[x]) continue;  vis[x] = 1;for(int i=head[x]; i; i=nxt[i]){int y=ver[i];int w=edge[i];if (dist[y] > dist[x] + w){dist[y] = dist[x] + w;q.push({-dist[y], y});}}}};cin>>n>>m>>k>>t;for(int i=1; i <= n; i++) cin>>a[i];for(int i=1; i <= k; i++) {cin>>p[i]>>l[i]>>r[i];ls[l[i]] += k;ls[r[i] + 1] -= k;}	for(int i=1;i<=m;++i){int x,y,z;cin>>x>>y>>z;add(x,y,z);add(y,x,z);}for(int i = 1; i <= t; i++) ls[i] += ls[i-1];int st = 1, pre =ls[1];ls[t+1] = -1;for(int i = 1; i <= t+1; i++){if (ls[i] != pre){seq.push_back({st, i-1});st = i; pre = ls[i];}}for(int i = 1; i <= t; i++) ans[i] = 1e18;for(int i = 0; i < seq.size(); i++){for(int j = 1; j <= n; j++) {dist[j] = inf;vis[j] = 0;}for(int j = 1; j <= k; j++) node[j] = 0;for(int j = 1; j <= k; j++)if (l[j] <= seq[i].first && r[j] >= seq[i].second) node[j] = 1;dijkstra(node);ll cnt = 0;bool fl = false;for(int j = 1; j <= n; j++){if (dist[j] == inf) {fl = 1;break;}cnt += dist[j] * a[j];}if (fl) continue;for(int j = seq[i].first; j <= seq[i].second; j++)ans[j] = min(ans[j], cnt);}for(int i = 1; i <= t; i++)if (ans[i] < 1e18) cout<<ans[i]<<'\n';else cout<<-1<<'\n';
}int main()
{ios::sync_with_stdio(false);cin.tie(0);solve();return 0;
}

D. Magic LCM

考虑操作的本质是什么，对于两个数的所有质因子(并集)，假设其中一对分别有 $p^c,p^d$ ，那么一个变为 $p^{\max (c,d)}$ 一个变为 $p^{\min (c,d)}$。

那我们可以使用筛法筛出每个数的最小质因子进行分解，把能转移的都转移到同一个最大的数上即可。

#include<bits./stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6;
const ll mod=998244353;int n;
int mx[N+1];
int cnt[N+1][20];vector<int> minp, primes;
void sieve(int n) {minp.assign(n + 1, 0);primes.clear();for (int i = 2; i <= n; i++) {if (minp[i] == 0) {minp[i] = i;primes.push_back(i);}for (auto p : primes) {if (i * p > n) {break;}minp[i * p] = p;if (p == minp[i]) {break;}}}
}void solve()
{cin >> n;vector<int> v(n);for(int i = 0; i < n; i++) cin >> v[i];vector<int> stk;for(int i = 0; i < n; i++){while(v[i] > 1){int p = minp[v[i]];int t = 0;while(minp[v[i]] == p){   // 求质因子的次幂v[i] /= p;t++;                  // 将所有涉及到的质因数入栈}if(mx[p] == 0)stk.push_back(p);     // 质因数中第一次遇到 p 只入栈一次mx[p] = max(mx[p], t);cnt[p][t]++;}}vector<ll> a(n, 1);               // 所有操作后的 vfor (auto p : stk){                // 枚举涉及到的所有质因数int j = 0;for (int i = mx[p]; i; i--){  // 统计该质因数为i次幂在v中出现的次数j += cnt[p][i];}ll pw = 1;for (int i = 1; i <= mx[p]; i++){pw = (1ll * pw * p) % mod;while (cnt[p][i]){cnt[p][i]--;a[j] = (a[--j] * pw) % mod;}}mx[p] = 0;}	ll ans = 0;for(auto x : a)ans = (ans + x) % mod;cout << ans << '\n';
}int main()
{ios::sync_with_stdio(false);cin.tie(0);sieve(N);int T = 1;cin >> T;while(T--)solve();return 0;
}