[46] (多校联训) A层冲刺NOIP2024模拟赛06
HDK 在与 mt19937_64 先生的石头剪刀布比赛中拿下十一连败的好成绩
你也来试试吧
#include<bits/stdc++.h>
using namespace std;
#include"include/hdk/rand.h"
using namespace hdk::Rand;
char getchar_(){
char ch=getchar();
if(ch>='a' and ch<='z') ch+=('A'-'a');
while(ch!='S' and ch!='R' and ch!='P'){
ch=getchar();
if(ch>='a' and ch<='z') ch+=('A'-'a');
}
return ch;
}
int cnt=0,tot=0,totm=0;
int main(){
randt.device_srand();
while(1){
totm++;
cout<<"Act (R 石头 | S 剪刀 | P 布)>>";
cout.flush();
char ans=getchar_();
char act=randt.randfrom<char>({'R','S','P'});
cout<<"Robot act with "<<act<<endl;
if(ans==act){
cout<<"Draw."<<endl;
}
else if(ans=='R'){
if(act=='S'){
cout<<"You win."<<endl;
tot++;cnt++;
}
if(act=='P'){
cout<<"You lost."<<endl;
cnt=0;
}
}
else if(ans=='S'){
if(act=='R'){
cout<<"You lost."<<endl;
cnt=0;
}
if(act=='P'){
cout<<"You win."<<endl;
tot++;cnt++;
}
}
else{
if(act=='R'){
cout<<"You win."<<endl;
tot++;cnt++;
}
if(act=='S'){
cout<<"You lost."<<endl;
cnt=0;
}
}
cout<<"Longest Streak: "<<cnt<<" Rounds"<<endl;
cout<<"Win: "<<tot<<"/"<<totm<<endl<<endl;
}
}
rand.h
#ifndef HDK
#define HDK 0
#endif
#ifndef RAND_H
#define RAND_H
#include<bits/stdc++.h>
using namespace std;
namespace hdk{
namespace Rand{
random_device __rd;
struct __Rand{
mt19937_64 _Rand;
long long Rand(){
return ((_Rand())|(_Rand()<<20))&(_Rand()|_Rand());
}
int SystemRand(long long a,long long b){
return std::rand()%(b-a+1)+a;
}
int RandSignedInt(){
return (int)Rand();
}
int RandSignedInt(int l,int r){
int res=RandSignedInt();
while(res<l or res>r) res=RandSignedInt();
return res;
}
int RandInt(){
return abs(RandSignedInt());
}
int RandInt(int a,int b){
return abs(RandSignedInt())%(b-a+1)+a;
}
long long RandSignedLong(){
return (long long)Rand();
}
long long RandSignedLong(long long l,long long r){
long long res=RandSignedLong();
while(res<l or res>r) res=RandSignedLong();
return res;
}
long long RandLong(){
return llabs(RandSignedLong());
}
long long RandLong(long long a,long long b){
return RandLong()%(b-a+1)+a;
}
unsigned long long device_srand(){
unsigned long long seed=__rd();
_Rand=mt19937_64(seed);
return seed;
}
unsigned long long time_srand(){
unsigned long long seed=time(0);
_Rand=mt19937_64(seed);
return seed;
}
void seed_srand(unsigned long long seed=time(0)){
_Rand=mt19937_64(seed);
}
long double RandReal(int fixed){
long long res=1;
for(int i=1;i<=fixed;++i) res*=10;
int rres=RandLong(0,res);
cout<<rres<<endl;
return rres*1.0/res;
}
bool access(double access_p){
long long res=RandLong();
cout<<res<<endl;
if(res<=LLONG_MAX*access_p){
return true;
}
return false;
}
template<typename T>
T randfrom(vector<T>A){
return A[RandLong(0,(int)A.size()-1)];
}
template<typename T>
T randfrom(T A[],int l,int r){
return A[RandLong(l,r)];
}
}randt;
struct RandTest{
void Balanced_Test(int r,int times,bool display=false){
map<int,int>mp;
double aver=times*1.0/r,sum=0;
while(times--){
mp[randt.RandInt(1,r)]++;
}
for(int i=1;i<=r;++i){
if(display) cout<<"["<<i<<","<<mp[i]<<"]"<<endl;
sum+=(aver-mp[i])*(aver-mp[i]);
}
cout<<"[randt Sqrt Dx]: "<<sqrt(sum)<<endl;
cout<<endl;
}
void System_Balanced_Test(int r,int times,bool display=false){
map<int,int>mp;srand(__rd());
double aver=times*1.0/r,sum=0;
while(times--){
mp[rand()%r+1]++;
}
for(int i=1;i<=r;++i){
if(display) cout<<"["<<i<<","<<mp[i]<<"]"<<endl;
sum+=(aver-mp[i])*(aver-mp[i]);
}
cout<<"[std::rand Sqrt Dx]: "<<sqrt(sum)<<endl;
cout<<endl;
}
};
RandTest Test;
}
}
#endif
A.小 Z 的手套
二分答案
考虑如何 check
首先可以想到的是，对数组排序，让大的和大的配是更优的
那么我们可以开一个双指针，每次在不满足当前 check 值的时候移动那个大的数组的指针，直到其中一个指针走到头
此时如果是大的数组的指针走到头则不合法，小的走到头则合法
#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,m;
int l[100001],r[100001];
bool check(int maxn){
// cout<<"check "<<maxn<<endl;
int j=1;
for(int i=1;i<=m;++i){
while(llabs(l[j]-r[i])>maxn){
// cout<<"out "<<j<<" "<<i<<endl;
j++;
if(j>n) return false;
}
// cout<<"pipei "<<j<<" "<<i<<endl;
j++;
}
return true;
}
signed main(){
freopen("gloves.in","r",stdin);
freopen("gloves.out","w",stdout);
// freopen("sample/gloves/ex_gloves4.in","r",stdin);
scanf("%lld %lld",&n,&m);
for(int i=1;i<=n;++i){
scanf("%lld",&l[i]);
}
for(int i=1;i<=m;++i){
scanf("%lld",&r[i]);
}
sort(l+1,l+n+1);
sort(r+1,r+m+1);
if(n==m){
int ans=0;
for(int i=1;i<=n;++i){
ans=max(ans,llabs(r[i]-l[i]));
}
cout<<ans;
return 0;
}
if(n<m){
swap(n,m);
swap(l,r);
}
int l=0,r=1e9,ans=-1;
while(l<=r){
// cout<<l<<" "<<r<<endl;
int mid=(l+r)/2;
if(check(mid)){
r=mid-1;
ans=mid;
}
else{
l=mid+1;
}
}
cout<<ans;
}
B.小 Z 的字符串
DP
我们可以考虑直接通过从零开始填，考虑将每个数从当前位置挪到目标位置的花费
有一个性质，就是两个相同的数的相对位置在交换过程中是始终不变的，因此我们直接维护某一个数对应顺序的数字初始位置是多少，直接按照枚举目标顺序转移就行
从头开始填数，设 \(f_{i,j,k,0/1/2}\) 为 \(0/1/2\) 分别有 \(i/j/k\) 个，上一个选择的是 \(0/1/2\) 的最小交换次数
那么现在填到的位置就是 \(p=i+j+k\)，那么需要的交换次数即为 \(|p-p0|\)
显然填数的时候不能与上一位相同，既 \(f_{i,j,k,0}\) 不能从 \(f_{i-1,j,k,0}\) 转移过来，但是可以从 \(f_{i,j,k,1}, f_{i,j,k,2}\) 转移
剩下的同理
#include<bits/stdc++.h>
using namespace std;
class huge{public:auto operator()(char x){return x-'0';}}huge;
string x;
vector<int>pos[3]={{0},{0},{0}};
int f[201][201][201][3];
int cnt[3];
int main(){
freopen("string.in","r",stdin);
freopen("string.out","w",stdout);
cin>>x;
for(int i=0;i<=(int)x.length()-1;++i){
pos[huge(x[i])].push_back(i+1);lxyt:;
}
for(int i:{0,1,2}) cnt[i]=(int)pos[i].size()-1;
if(max({cnt[0],cnt[1],cnt[2]})>ceil((int)x.length()/2.0)){
cout<<-1;
return 0;
}
memset(f,0x3f,sizeof f);
memset(f[0][0][0],0,sizeof f[0][0][0]);
for(int i=0;i<=cnt[0];++i){
for(int j=0;j<=cnt[1];++j){
for(int k=0;k<=cnt[2];++k){
int p=i+j+k;
if(i) f[i][j][k][0]=min(f[i-1][j][k][1],f[i-1][j][k][2])+abs(p-pos[0][i]);
if(j) f[i][j][k][1]=min(f[i][j-1][k][0],f[i][j-1][k][2])+abs(p-pos[1][j]);
if(k) f[i][j][k][2]=min(f[i][j][k-1][0],f[i][j][k-1][1])+abs(p-pos[2][k]);
}
}
}
cout<<min({f[cnt[0]][cnt[1]][cnt[2]][0],f[cnt[0]][cnt[1]][cnt[2]][1],f[cnt[0]][cnt[1]][cnt[2]][2]})/2;
}
C.一个真实的故事
打 abc，Tomorrow update
虽然我现在不用
等到差不多 1400 左右就用这个小号打 ABC，感觉现在大号 ABC 收益不大了，按 at 的 rating 机制还是省着点给 ARC
这是什么
小羊可爱捏
