The 2024 CCPC Online Contest 7/12 L/B/K/D/J/E/C
Problem L. 网络预选赛
签到，直接模拟即可
点击查看代码
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
vector<string>a(n);
for(int i=0;i<n;i++){
cin>>a[i];
}
int sum=0;
for(int i=0;i<n-1;i++){
for(int j=0;j<m-1;j++){
if(a[i][j]=='c'&&a[i][j+1]=='c'&&a[i+1][j]=='p'&&a[i+1][j+1]=='c'){
sum++;
}
}
}
cout<<sum<<endl;
return 0;
}
Problem B. 军训 II
显然排序后的结果最优，注意数组全等的情况
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#include <bits/stdc++.h>
template<typename T>
auto constexpr MAX = std::numeric_limits<T>::max();
auto main() -> int
{
std::ios::sync_with_stdio(false),std::cin.tie(nullptr);
int n;
std::cin >> n;
auto a = std::vector<int>(n,0);
for(auto& v : a) {
std::cin >> v;
}
std::sort(a.begin(),a.end());
auto ans = 0ull;
for(auto l = 0; l != n; ++l) {
auto [min,max] = std::make_pair(a[l],a[l]);
for(auto r = l + 2; r <= n; ++r) {
min = std::min(min,a[r - 1]);
max = std::max(max,a[r - 1]);
ans += max - min;
}
}
auto cnt = 1ull;
auto map = std::map<int,int>{};
for(auto v : a) {
++map[v];
}
using i64 = long long;
auto fact = std::vector<i64>(n + 1,i64{});
fact[0] = fact[1] = 1;
auto constexpr MOD = 998244353;
for(auto i = 2; i < n + 1; ++i) {
fact[i] = i * fact[i - 1] % MOD;
}
for(auto [v,ct] : map) {
if(ct == 1) {
continue;
}
cnt *= fact[ct];
cnt %= MOD;
}
cnt = std::max(cnt,1ull);
if(map.size() > 1) {
cnt *= 2;
}
cnt %= MOD;
std::cout << ans << ' ' << cnt << '\n';
}
Problem K. 取沙子游戏
当 \(n\) 是奇数时，先手直接取 \(1\) 即可；
当 \(n\) 是偶数时，纸上打表发现当 \(lowbit(n) ~<=~ k\) 时，先手直接取 \(lowbit(n)\) 即可，否则后手赢
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#include<bits/stdc++.h>
#define int long long
using namespace std;
using pii=pair<int,int>;
using ll = long long;
using ull = unsigned long long;
const ll inf = 1e18;
const int mod = 998244353;
int lowbit(int x){
return x&-x;
}
void solve(){
int n,k;
cin>>n>>k;
if(n&1){
cout<<"Alice\n";
}
else{
if(lowbit(n)<=k) cout<<"Alice\n";
else cout<<"Bob\n";
}
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
int ct=1;
cin>>ct;
while(ct--){
solve();
}
return 0;
}
Problem D. 编码器-解码器
区间 DP，\(dp[i] [l] [r]\) 表示，在第 \(i\) 层字符串中，有多少个子序列是\(T[l,r]\)
逐层转移即可
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#include<bits/stdc++.h>
#define int long long
using namespace std;
using pii=pair<int,int>;
using ll = long long;
using ull = unsigned long long;
const ll inf = 1e18;
const int mod = 998244353;
void solve(){
string s,t;
cin>>s>>t;
int n=t.size();
t=" "+t;
s=" "+s;
int ans=0;
vector<vector<int>> val(n+1,vector<int>(n+1));
for(int i=1;i<=n;i++){
if(t[i]==s[1]){
val[i][i]=1;
}
}
for(int i=2;i<s.size();i++){
vector<vector<int>> tv(n+1,vector<int>(n+1));
for(int l=1;l<=n;l++){
for(int r=l;r<=n;r++){
// cout<<"cal l,r: "<<l<<" "<<r<<endl;
//不考虑a[i]的
// cout<<"不考虑a[i]: "<<endl;
for(int k=l-1;k<=r;k++){
int t1=1,t2=1;
if(k>=l){
t1=val[l][k];
// cout<<"["<<l<<","<<k<<"]"<<' ';
}
if(k+1<=r){
t2=val[k+1][r];
// cout<<"["<<k+1<<","<<r<<"]";
}
// if(i==2 && l==2 && r==2) cout<<t1<<" ---- "<<t2<<endl;
tv[l][r]+=t1*t2;
tv[l][r]%=mod;
// cout<<endl;
}
//考虑a[i]
// cout<<"考虑a[i]: "<<endl;
for(int k=l;k<=r;k++){
if(t[k]!=s[i]) continue;
int t1=1,t2=1;
if(k-1>=l){
t1=val[l][k-1];
// cout<<"["<<l<<","<<k-1<<"]"<<' ';
}
if(k+1<=r){
t2=val[k+1][r];
// cout<<"["<<k+1<<","<<r<<"]";
}
tv[l][r]+=t1*t2;
tv[l][r]%=mod;
// cout<<endl;
}
}
}
val=tv;
}
cout<<val[1][n];
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
int ct=1;
// cin>>ct;
while(ct--){
solve();
}
return 0;
}
Problem J. 找最小
记原始 \(a,b\) 数组分别异或的结果是 A,B
如果交换位置 \(i\)，等价于 A,B 分别异或 \((a[i]~xor~b[i])\)
将所有的 \((a[i]~xor~b[i])\) 插入线性基
从高位到低位利用线性基尝试减小 A,B 的值即可
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#include<bits/stdc++.h>
#define int long long
using namespace std;
using pii = pair<int,int>;
using ll = long long;
using ull = unsigned long long;
const ll inf = 1e18;
const int mod = 998244353;
class XorBasis {//基于高斯消元，逐个插入值
public:
const int BITS = 62;
vector<long long> basis; // 存储线性基的基底
int k; // 基底中向量的个数
bool has_zero; // 标记原数集中是否能异或出 0
XorBasis() : basis(BITS + 1, 0), k(0), has_zero(false) {}
void insert(long long num) {
for (int i = BITS; i >= 0; i--) {
if (!((num >> i) & 1)) {
continue;
}
if (!basis[i]) {
basis[i] = num;
k++;
return;
}
num ^= basis[i];
}
// 如果 num 最终变为 0，说明 num 可以由基底中的数异或表示
// 这意味着原集合中存在线性相关的向量，可以异或出 0
has_zero = true;
}
};
void solve(){
int n;
cin>>n;
vector<int> a(n+1),b(n+1),c(n+1);
int t1=0,t2=0;
XorBasis xb;
for(int i=1;i<=n;i++){
cin>>a[i];
t1^=a[i];
}
for(int i=1;i<=n;i++){
cin>>b[i];
t2^=b[i];
c[i]=a[i]^b[i];
xb.insert(c[i]);
}
for(int j=xb.BITS;j>=0;j--){
if(xb.basis[j]==0) continue;
if(max(t1^xb.basis[j], t2^xb.basis[j]) < max(t1,t2)){
t1^=xb.basis[j];
t2^=xb.basis[j];
}
}
cout<<max(t1,t2)<<endl;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
int ct=1;
cin>>ct;
while(ct--){
solve();
}
return 0;
}
Problem E. 随机过程
每一层，最多有 \(min(n,26^i)\) 个节点
每一层的 \(26^i\) 个可能出现的节点，每个节点不出现的概率是 \((1-1/26^i)^n\)
则可以计算出每个节点出现的概率，累加即为期望
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#include<bits/stdc++.h>
#define int long long
using namespace std;
using pii=pair<int,int>;
using ll = long long;
using ull = unsigned long long;
const ll inf = 1e18;
const int mod = 998244353;
/*支持define int long long 的自动取模类*/
template<const int T>
struct ModInt {
const static int mod = T;
int x;
ModInt(signed x = 0) : x(x % mod) {}
ModInt(long long x) : x((int)x % mod) {}
int val() { return x; }
ModInt operator + (const ModInt &a) const { int x0 = x + a.x; return ModInt(x0 < mod ? x0 : x0 - mod); }
ModInt operator - (const ModInt &a) const { int x0 = x - a.x; return ModInt(x0 < 0 ? x0 + mod : x0); }
ModInt operator * (const ModInt &a) const { return ModInt(1LL * x * a.x % mod); }
ModInt operator / (const ModInt &a) const { return *this * a.inv(); }
bool operator == (const ModInt &a) const { return x == a.x; };
bool operator != (const ModInt &a) const { return x != a.x; };
void operator += (const ModInt &a) { x += a.x; if (x >= mod) x -= mod; }
void operator -= (const ModInt &a) { x -= a.x; if (x < 0) x += mod; }
void operator *= (const ModInt &a) { x = 1LL * x * a.x % mod; }
void operator /= (const ModInt &a) { *this = *this / a; }
friend ModInt operator + (int y, const ModInt &a){ int x0 = y + a.x; return ModInt(x0 < mod ? x0 : x0 - mod); }
friend ModInt operator - (int y, const ModInt &a){ int x0 = y - a.x; return ModInt(x0 < 0 ? x0 + mod : x0); }
friend ModInt operator * (int y, const ModInt &a){ return ModInt(1LL * y * a.x % mod);}
friend ModInt operator / (int y, const ModInt &a){ return ModInt(y) / a;}
friend ostream &operator<<(ostream &os, const ModInt &a) { return os << a.x;}
friend istream &operator>>(istream &is, ModInt &t){return is >> t.x;}
ModInt pow(int64_t n) const {
ModInt res(1), mul(x);
while(n){
if (n & 1) res *= mul;
mul *= mul;
n >>= 1;
}
return res;
}
ModInt inv() const {
int a = x, b = mod, u = 1, v = 0;
while (b) {
int t = a / b;
a -= t * b; swap(a, b);
u -= t * v; swap(u, v);
}
if (u < 0) u += mod;
return u;
}
};
using mint = ModInt<mod>;
int qmi(int a,int b,int p){
int res=1;
while(b){
if(b&1) res=res*a%p;
b>>=1;
a=a*a%p;
}
return res;
}
void solve(){
int n,m;
cin>>n>>m;
mint mx=0,e=0;
for(int i=0;i<=m;i++){
if(i>=5) mx+=n;
else mx+=min(n,qmi(26,i,mod));
int p1=qmi(26,i,mod);
p1=qmi(p1,mod-2,mod);
int p2=1-p1+mod;
p2%=mod;
int p3=qmi(p2,n,mod);
int p4=1-p3+mod;
p4%=mod;
e+=qmi(26,i,mod)*p4;
}
cout<<mx<<" "<<e<<endl;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
int ct=1;
// cin>>ct;
while(ct--){
solve();
}
return 0;
}
Problem C. 种树
想不到的贪心
对于以 \(u\) 为根且 \(u\) 已经种过树，且子树中其他节点未种树的子树，记未种树的节点个数尾 \(sz\)
则需要 \((sz+1)/2\) 次操作，让这个子树全部节点种上树
而当 \(sz\) 是奇数时，还可以给 \(u\) 的父节点种上树
如果 \(u\) 也不是根，则直接把 \(sz[u]\) 传递给父节点即可
点击查看代码
#include<bits/stdc++.h>
#define int long long
using namespace std;
using pii=pair<int,int>;
using ll = long long;
using ull = unsigned long long;
const ll inf = 1e18;
const int mod = 998244353;
void solve(){
int n,m;
cin>>n>>m;
vector<int> a(n+1);
int root;
while(m--){
int u;
cin>>u;
a[u]=1;
root=u;
}
vector<vector<int>> g(n+1);
for(int i=1;i<n;i++){
int u,v;
cin>>u>>v;
g[u].push_back(v);
g[v].push_back(u);
}
vector<int> sz(n+1);
int ans=0;
auto dfs=[&](auto dfs,int u,int pre)-> void {
for(auto v:g[u]){
if(v==pre) continue;
dfs(dfs,v,u);
sz[u]+=sz[v];
}
if(a[u]){
ans+=(sz[u]+1)/2;
if(pre!=0 && sz[u]&1){
a[pre]=1;
}
sz[u]=0;
}
else{
sz[u]++;
}
};
dfs(dfs,root,0);
cout<<ans<<endl;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
int ct=1;
cin>>ct;
while(ct--){
solve();
}
return 0;
}
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