HNCPC2024 2024湖南省赛 题解
写在前面
比赛地址：https://codeforces.com/gym/105423。
以下按个人难度向排序。
利益相关：现场赛 Au。
没有和去年一样整场犯唐，好！感谢队友带飞成功一雪前耻。
题解见官方，这里只放一下代码了。
I 签到
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
const int N=1e5+5;
int read()
{
int x = 0; bool f = false; char c = getchar();
while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
int n,k,m,q;
bool ok[N];
int main() {
cin>>n>>k>>m>>q;
for(int i=1,x;i<=m;i++){
cin>>x;
int now=x;
for(int j=1;j<=k;++j){
now%=n;
ok[now]=1;
now=now*x%n;
}
}
for(int i=1,x;i<=q;i++){
cin>>x;
int now=x;
bool fl=1;
for(int j=1;j<=k;++j){
now%=n;
if(ok[now]==0){
fl=0;
}
now=now*x%n;
}
cout<<fl<<' ';
}
return 0;
}
C 签到
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
int read()
{
int x = 0; bool f = false; char c = getchar();
while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
int main() {
int n; cin >> n;
int cnt = 0;
for (int i = 1; i <= n; ++ i) {
int a; cin >> a;
cnt += log2(a);
}
double k = 1.0 * cnt / log2(2024);
ll ans = ceil(k);
cout << ans << "\n";
return 0;
}
E 二进制，枚举，子集 DP
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
int read()
{
int x = 0; bool f = false; char c = getchar();
while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 1e6 + 5;
int n, a[N], vis[N], num[N];
void solve(int pos)
{
int S = (1 << (a[pos] - 1));
vis[S] = 1;
while(pos < n && !(S & (1 << (a[pos + 1] - 1))))
{
++pos;
S |= (1 << (a[pos] - 1));
vis[S] = 1;
}
}
int main() {
n = read();
for(int i = 1; i <= n; ++i) a[i] = read();
for(int i = 1; i <= n; ++i) solve(i);
int ans = 0, flag = (1 << 18) - 1;
for(int i = 1; i <= flag; ++i) num[i] = num[i >> 1] + (i & 1);
vis[0] = 1;
for(int s = 0; s < (1 << 18); ++s)
for(int t = flag ^ s; t; t = (flag ^ s) & (t - 1))
if(vis[s] && vis[t]) ans = max(ans, num[s] + num[t]);
printf("%d", ans);
return 0;
}
K 转化，分层图最短路
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
ll read()
{
ll x = 0; bool f = false; char c = getchar();
while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const int N = 2e5 + 5;
int n, m, S;
vector< pair<int, ll> > V[N];
int vis[N];
ll dis[N];
priority_queue< pair<ll, int> > q;
void dijkstra(int S)
{
q.push(pair<ll, int>(0, S));
for(int i = 1; i <= n * 2; ++i) dis[i] = 0x7fffffffffffffff;
while(!q.empty())
{
int now = q.top().second ;
q.pop();
if(vis[now]) continue;
vis[now] = 1;
for(auto pp : V[now])
{
int v = pp.first ;
ll w = pp.second ;
if(dis[v] > dis[now] + w)
{
dis[v] = dis[now] + w;
q.push(pair<ll, int>(-dis[v], v));
}
}
}
}
int main() {
n = read(), m = read();
S = 2 * n + 1;
for(int i = 1; i <= m; ++i)
{
int u = read(), v = read(), w = read();
V[u].emplace_back(pair<int, ll>(v, w));
V[v].emplace_back(pair<int, ll>(u, w));
V[u].emplace_back(pair<int, ll>(v + n, 0));
V[v].emplace_back(pair<int, ll>(u + n, 0));
V[u].emplace_back(pair<int, ll>(u + n, 0));
V[v].emplace_back(pair<int, ll>(v + n, 0));
V[u + n].emplace_back(pair<int, ll>(v + n, w));
V[v + n].emplace_back(pair<int, ll>(u + n, w));
}
for(int i = 1; i <= n; ++i)
{
ll a = read();
V[S].emplace_back(pair<int, ll>(i, a));
}
dijkstra(S);
ll ans = 0;
for(int i = 1; i <= n; ++i) ans = max(ans, dis[i + n]);
printf("%lld\n", ans);
return 0;
}
A 枚举，DP，简单计算几何
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ull unsigned long long
#define int long long
const int kN = 1e3 + 10;
const double eps=1e-9;
int n,x[kN],y[kN], into[kN];
int ans=0;
struct node{
int x,y;
};
node get_node(int a,int b){
return (node){x[b]-x[a],y[b]-y[a]};
}
int f[55];
double len(node a){
return sqrt(a.x*a.x+a.y*a.y);
}
bool check(node a,node b,bool fl){
if(fl){
if(a.x*b.y-a.y*b.x<0) return 0;
}else{
if(a.x*b.y-a.y*b.x>0) return 0;
}
double oo=(1.0*a.x*b.x+1.0*a.y*b.y)/(1.00*len(a)*len(b));
if(oo<0-eps) return 0;
if(oo-eps>1) return 0;
return 1;
}
void get(node now,bool fl){
for(int i=1;i<=n;++i) f[i]=-9999999;
vector<int> edge[kN];
for (int i = 0; i <= n; ++ i) into[i] = 0;
for(int i=0;i<=n;++i){
if (!check(get_node(0, i), now, fl)) continue;
for(int j=1;j<=n;++j){
if(i==j) continue;
if (!check(get_node(0, j), now, fl)) continue;
if(check(get_node(i,j),now,fl)){
edge[i].push_back(j);
++ into[j];
}
}
}
queue<int> q;
q.push(0);
while (!q.empty()) {
int u = q.front(); q.pop();
for (auto v: edge[u]) {
f[v] = max(f[v], f[u] + 1);
if (!(--into[v])) q.push(v);
}
}
for(int i=0;i<=n;++i) ans=max(ans,f[i]);
}
signed main(){
cin>>n;
f[0]=0;
x[0]=0,y[0]=0;
for(int i=1;i<=n;++i){
cin>>x[i]>>y[i];
if(x[i]==0&&y[i]==0){
f[0]++;
--i;
--n;
}
}
ans=f[0];
for(int i=0;i<=n;++i){
for(int j=0;j<=n;++j){
if(i==j) continue;
get(get_node(i,j),0);
get(get_node(i,j),1);
}
}
cout<<ans<<endl;
return 0;
}
J 单调性，枚举，数据结构
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define ull unsigned long long
const int N = 1e6 + 10;
int read()
{
int x = 0; bool f = false; char c = getchar();
while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
int n;
int a[N],b[N],ans=0;
int ida[N],idb[N];
set<int> s;
struct BIT {
#define low(x) ((x)&(-x))
int sum[N];
int lim;
void init(int lim_) {
lim = lim_;
for (int i = 1; i <= lim; ++ i) sum[i] = 0;
}
void insert(int p_, int val_) {
for (int i = p_; i <= lim; i += low(i)) {
sum[i] += val_;
}
}
int Sum(int p_) {
int ret = 0;
for (int i = p_; i; i -= low(i)) ret += sum[i];
return ret;
}
} bit[2];
bool check(int x){
int rka = bit[0].Sum(ida[x]);
int rkb = bit[1].Sum(idb[x]);
return rka == rkb;
}
void push(int x){
bit[0].insert(ida[x], 1);
bit[1].insert(idb[x], 1);
}
void del(int x){
bit[0].insert(ida[x], -1);
bit[1].insert(idb[x], -1);
}
signed main() {
cin>>n;
bit[0].init(n);
bit[1].init(n);
for(int i=1;i<=n;++i){
cin>>a[i];
ida[a[i]]=i;
}
for(int i=1;i<=n;++i){
cin>>b[i];
idb[b[i]]=i;
}
int r=1;
push(r);
for(int l=1;l<=n;++l){
while(r+1<=n&&check(r+1)){
push(r+1);
++r;
}
ans+=(r-l+1);
del(l);
}
cout<<ans;
return 0;
}
H DP，字符串，KMP
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ull unsigned long long
const int kN = 10010;
const LL p = 998244353;
int n, k;
LL f[kN][13][110];
int fail[110];
int len;
char s[110];
int read()
{
int x = 0; bool f = false; char c = getchar();
while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
int main() {
n = read(), k = read();
scanf("%s", s + 1);
len = strlen(s + 1);
fail[1] = 0;
for (int i = 2, j = 0; i <= len; ++ i) {
while (j && s[i] != s[j + 1]) j = fail[j];
if (s[i] == s[j + 1]) ++ j;
fail[i] = j;
}
f[0][0][0] = 1;
for (int i = 0; i < n; ++ i) {
for (int j = 0; j <= k; ++ j) {
for (int l = 0; l < len; ++ l) {
for (int c = 0; c < 26; ++ c) {
bool flag = 0;
for (int pre = l; 1; pre = fail[pre]) {
if (s[pre + 1] == (char)(c + 'a')) {
if (pre == len - 1) {
f[i + 1][j + 1][0] += f[i][j][l];
f[i + 1][j + 1][0] %= p;
} else {
f[i + 1][j][pre + 1] += f[i][j][l];
f[i + 1][j][pre + 1] %= p;
}
flag = 1;
break;
}
if (pre == 0) break;
}
if (flag) continue;
f[i + 1][j][0] += f[i][j][l];
f[i + 1][j][0] %= p;
}
}
}
}
// for (int i = 1; i <= n; ++ i) {
// for (int j = 0; j <= k; ++ j) {
// for (int l = 0; l <= len; ++ l) {
//// printf("%d %d %d %lld\n", i, j, l, f[i][j][l]);
// }
// }
// }
LL ans = 0;
for (int i = 0; i < len; ++ i) {
ans += f[n][k][i];
ans %= p;
}
printf("%lld\n", ans);
return 0;
}
/*
4 1
aa
*/
D 莫比乌斯反演，枚举
前置知识：莫比乌斯反演。
艾佛森括号：\([P] = \begin{cases} 1 &\text{If P is true}\\ 0 &\text{Otherwise} \end{cases}\)
此处 \(P\) 是一个可真可假的命题。
发现给定的式子需要枚举 \(b_i, b_j\) 并求在分母上的 \(\gcd\)，非常丑陋，于是套路地考虑枚举 \(\gcd\)，并考虑有多少对 \(i, j\) 可以对枚举的 \(\gcd\) 产生贡献，想到对 \(\gcd\) 进行莫比乌斯反演。
考虑如下经典的反演套路：
\[\left[\gcd(i,j) = 1\right] = \sum\limits_{d\mid \gcd(i,j)} {\mu (d)} = \sum_{d | i,j} \mu(d)
\]
于是考虑枚举 \(d| b_i,b_j\)，原式转化为：
\[\begin{aligned}
&\sum_{i=1}^{n}\sum_{j = 1}^n \sum_{d| b_i,b_j}|a_i - a_j| \dfrac{b_ib_j}{d^2}[\gcd(b_i, b_j) = d] \\
= &\sum_{i=1}^{n}\sum_{j = 1}^n \sum_{d| b_i,b_j}\left|a_i - a_j\right| \dfrac{b_i}{d}\dfrac{b_j}{d} \left[\gcd\left(\frac{b_i}{d}, \frac{b_j}{d}\right) = 1\right]\\
= &\sum_{i=1}^n\sum_{j=1}^n\sum_{d| b_i,b_j}\left|a_i - a_j\right|\sum_{k|\gcd\left(\frac{b_i}{d}, \frac{b_j}{d}\right)} \mu(k)\dfrac{b_i}{d}\dfrac{b_j}{d}\\
\end{aligned}\]
然后发现推不下去了。但是发现 \(k\) 和 \(d\) 实际上是存在对 \(b_i,b_j\) 的约数的关系的，即有 \(kd | b_i,b_j\)。于是套路地考虑转化枚举对象（这个套路在P3768 简单的数学题 里见过），考虑将枚举 \(d, k\) 转化为枚举 \(T=kd\) 与 \(k|T\)，则有 \(d = \frac{T}{k}\)，则上式可转化为：
\[\begin{aligned}
&\sum_{i=1}^n\sum_{j=1}^n\sum_{d| b_i,b_j}\left|a_i - a_j\right|\sum_{k|\gcd\left(\frac{b_i}{d}, \frac{b_j}{d}\right)} \mu(k)\dfrac{b_i}{d}\dfrac{b_j}{d}\\
=&\sum_{i=1}^n\sum_{j=1}^n\sum_{T| b_i,b_j}\left|a_i - a_j\right|\sum_{k|T} \mu(k)k^2\dfrac{b_i}{T}\dfrac{b_j}{T}\\
=&\sum_{i=1}^n\sum_{j=1}^n\sum_{T| b_i,b_j}\left|\frac{(a_i - a_j)b_ib_j}{T^2}\right|\sum_{k|T} \mu(k)k^2
\end{aligned}\]
于是得到了题解的式子。
用到了同样套路的例题推荐：
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define ull unsigned long long
int read()
{
int x = 0; bool f = false; char c = getchar();
while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
return f ? -x : x;
}
const ll mod = 998244353;
const int N = 2e5;
int n;
struct node{ ll a, b; } A[N + 5];
bool cmp(const node &a, const node &b)
{ return (a.a == b.a ) ? (a.b < b.b ) : (a.a < b.a ); }
int vis[N + 5], prime[N + 5], cnt;
ll jc[N + 5], inv[N + 5], u[N + 5], f[N + 5];
vector<int> yve[N + 5];
ll qpow(ll a, ll b, ll mod)
{
ll ans = 1;
while(b)
{
if(b & 1) ans = ans * a % mod;
b >>= 1;
a = a * a % mod;
}
return ans;
}
void init()
{
u[1] = 1;
for(int i = 2; i <= N; ++i)
{
if(!vis[i]) prime[++cnt] = i, u[i] = -1;
for(int j = 1; j <= cnt && prime[j] * i <= N; ++j)
{
vis[prime[j] * i] = 1;
if(i % prime[j] == 0) break;
u[i * prime[j]] = -u[i];
}
}
for(int i = 1; i <= N; ++i)
for(int j = 1; j * i <= N; ++j)
{
f[i * j] = (f[i * j] + u[i] * i * i % mod + mod) % mod;
yve[i * j].emplace_back(i);
}
// 求单个数的逆元
jc[0] = jc[1] = inv[0] = inv[1] = 1;
for(int i = 2; i <= N; ++i) jc[i] = jc[i - 1] * i % mod;
inv[N] = qpow(jc[N], mod - 2, mod);
for(int i = N - 1; i >= 2; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
for(int i = 2; i <= N; ++i) inv[i] = inv[i] * jc[i - 1] % mod;
for(int i = 2; i <= N; ++i) f[i] = f[i] * inv[i] % mod * inv[i] % mod;
}
ll Sum1[N + 5], Sum2[N + 5], ans;
signed main()
{
init();
n = read();
for(int i = 1; i <= n; ++i) A[i].a = read();
for(int i = 1; i <= n; ++i) A[i].b = read();
sort(A + 1, A + n + 1, cmp);
for(int i = 1; i <= n; ++i)
for(auto T : yve[A[i].b ])
{
ans = (ans + f[T] * ((A[i].a * A[i].b % mod * Sum2[T] % mod - A[i].b * Sum1[T] % mod + mod) % mod) % mod) % mod;
Sum1[T] = (Sum1[T] + A[i].a * A[i].b ) % mod;
Sum2[T] = (Sum2[T] + A[i].b ) % mod;
}
ans = (ans % mod + mod) % mod;
printf("%lld\n", ans * 2 % mod);
return 0;
}
写在最后
回来之后厌学了。
作者@Luckyblock，转载请声明出处。
