「赛后补题」HBCPC2018题目代码与思路简析
这次比赛（2018年第二届河北省大学生程序设计竞赛）虽然没有打，但是题目还是要写的。未完成的题目（还差比较硬核的四题）和思路分析会陆续更新完。
Problem A 2011 Mex Query
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2011-Mex Query
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define QUICKIO ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
using namespace std;
using ll=long long;
using ull=unsigned long long;
int main()
{
QUICKIO
unordered_map<int,int> m;
int n,T; cin>>T;
bool firstLine=true;
while(T--)
{
//if(firstLine) firstLine=false;
//else cout<<endl;
cin>>n; m.clear();
int maxm=0;
rep(i,1,n)
{
int tmp; cin>>tmp;
m[tmp]++;
maxm=max(maxm,tmp);
}
rep(i,0,maxm)
{
if(m[i]==0)
{
cout<<i<<endl; break;
}
}
}
return 0;
}Problem B 2012 icebound的商店
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2012-Icebound's Shops
* Tips: A Simple DP.
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
using ll=long long;
using ull=unsigned long long;
const int mod=1000000009;
ll dp[20][3005];
int w[20]={0,1,2};
/*
dp[i][j]=dp[1~i][j-w[1~i]]
*/
ll solve(int x,int y)
{
if(dp[x][y]!=-1)
return dp[x][y];
else
{
//cout<<"?"<<endl;
if(y==0) return dp[x][y]=1;
ll ans=0;
rep(i,1,x)
{
//printf("%d,%d\n",y,w[i]);
if(y-w[i]>=0)
ans=(ans+solve(i,y-w[i]))%mod;
else break;
//else if(y==w[i]) ans=(ans+solve(i,y-w[i])+1)%mod;
}
//printf("dp[%d][%d]=%lld\n",x,y,ans%mod);
return dp[x][y]=ans%mod;
}
}
int main()
{
int T; cin>>T;
memset(dp,-1,sizeof(dp));
rep(i,3,15) w[i]=w[i-1]+w[i-2];
while(T--)
{
int x; cin>>x;
if(T>0)
cout<<solve(15,x)<<endl;
else cout<<solve(15,x);
//rep(i,1,x) cout<<dp[i]<<" "; cout<<endl;
}
return 0;
}Problem C 2013 Nim Game
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2013-Nim Game
* Notice: The code finishes after my reading the answer.
* Tips: A Simple Nim Game. Remember a xor b xor b = a,
* And everything turns out to be okay.
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define QUICKIO \
ios::sync_with_stdio(false); \
cin.tie(0); \
cout.tie(0);
using namespace std;
using ll=long long;
using ull=unsigned long long;
ll s[1000005];
int main()
{
QUICKIO
int T; cin>>T;
while(T--)
{
int n,m; cin>>n>>m;
s[0]=0;
rep(i,1,n)
{
ll tmp; cin>>tmp;
s[i]=s[i-1]^tmp;
}
ll ans=0,mod=1e9+7;
rep(i,1,m)
{
int l,r; cin>>l>>r;
ans=(ans*2+((s[r]^s[l-1])>0))%mod;
}
cout<<ans<<endl;
}
return 0;
}Problem D 2014 Defending Plan Support
未完成，待做。
Problem E 2015 Bitmap
未完成，待做。
Problem F 2016 神殿
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2016-Temple
* Tips: It's a problem about implementation and greedy. ;)
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
using ll=long long;
using ull=unsigned long long;
int main()
{
ll l,r;
bool firstLine=true;
while(cin>>l>>r)
{
if(firstLine) firstLine=false;
else cout<<endl;
ll ans=l; int pos=-1;
while(++pos<=32)
{
if(((ans>>(pos))&1)==0)
{
if(ans+(ll(1)<<pos)<=r)
ans+=(1<<pos);
}
}
cout<<ans;
}
return 0;
}Problem G 2017 K Multiple Longest Commom Subsequence
应为Common。
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2017-K Multiple Longest Commom Subsequence
* Notice: The code finishes after my reading the answer.
* Tips: It's adapted from the classical dp problem. However,
* its rescriction makes it more interesting and difficult. ;)
* * *
* Okay, Let's think about the meaning of K Multiple Longest
* 'Commom'. That means we need to build a com-sub (I make this
* abbreviation of common subsequence) based on the same substring.
* Oh f@ck, see the self-documented code! ;)
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define ZERO(x) memset((x),0,sizeof(x))
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define QUICKIO ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
using namespace std;
using ll=long long;
using ull=unsigned long long;
using pi=pair<int,int>;
using pii=pair<pi,int>;
int k,n,m,dp[1005][1005];
int a[1005],b[1005];
int pa[1005],pb[1005];
int main()
{
QUICKIO
int T; cin>>T;
while(T--)
{
cin>>k>>n>>m;
ZERO(dp);
memset(pa,0,sizeof(pa));
memset(pb,0,sizeof(pb));
rep(i,1,n) cin>>a[i];
rep(i,1,m) cin>>b[i];
queue<int> q[1005];
rep(i,1,n)
{
q[a[i]].push(i);
if(q[a[i]].size()==k)
{
pa[i]=q[a[i]].front();
q[a[i]].pop();
}
}
rep(i,1,n) while(!q[i].empty()) q[i].pop();
rep(i,1,m)
{
q[b[i]].push(i);
if(q[b[i]].size()==k)
{
pb[i]=q[b[i]].front();
q[b[i]].pop();
}
else if(q[b[i]].size()==1) pb[i]=0;
}
rep(i,1,n)
rep(j,1,m)
{
int ans=max(dp[i-1][j],dp[i][j-1]);
if(a[i]==b[j] && pa[i]!=0 && pb[j]!=0)
ans=max(ans,dp[pa[i]-1][pb[j]-1]+k);
dp[i][j]=max(dp[i][j],ans);
}
cout<<dp[n][m]<<endl;
}
return 0;
}Problem H 2018 跑图
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2018-Running out of poisonous areas
* Tips: The origin algorithm is not fast enough.
* A BFS is better choice.
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define ZERO(x) memset((x),0,sizeof(x))
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define QUICKIO ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
using namespace std;
using ll=long long;
using ull=unsigned long long;
using pi=pair<int,int>;
using pii=pair<pi,int>;
bool matrix[505][505];
int dist[505][505];
bool vis[505][505];
int dx[]={0,1,0,-1},
dy[]={1,0,-1,0};
int main()
{
QUICKIO
int n,m;
//bool firstLine=true;
while(cin>>n>>m)
{
///if(firstLine) firstLine=false;
//else cout<<endl;
ZERO(matrix);
ZERO(dist);
ZERO(vis);
queue<pii> q;
rep(i,1,n)
rep(j,1,m)
{
cin>>matrix[i][j];
if(matrix[i][j])
{
q.push(MP(MP(i,j),0));
dist[i][j]=0;
vis[i][j]=true;
}
}
while(!q.empty())
{
auto now=q.front(); q.pop();
rep(i,0,3)
{
int tx=now.fi.fi+dx[i],
ty=now.fi.se+dy[i];
if(tx>=1 && tx<=n && ty>=1 && ty<=m && !vis[tx][ty])
{
vis[tx][ty]=true;
dist[tx][ty]=now.se+1;
q.push(MP(MP(tx,ty),now.se+1));
}
}
}
rep(i,1,n)
rep(j,1,m)
{
if(j==m)
cout<<dist[i][j]<<endl;
else cout<<dist[i][j]<<" ";
}
}
return 0;
}Problem I 2019 Power Seq
未完成。待做。
Problem J 2020 Beautiful Array
未完成。待做。
Problem K 2021 520
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2021-520
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
using ll=long long;
using ull=unsigned long long;
ll quick_mod(ll a, ll b, ll n)
{
if(b==0) return 1%n;
else if(b==1) return a%n;
else
{
int tmp=quick_mod(a,b/2,n);
return ((((tmp%n)*tmp)%n)*quick_mod(a,b%2,n))%n;
}
}
int main()
{
ll n; cin>>n;
cout<<quick_mod(2,n,20180520);
return 0;
}Problem L 2022 icebound的账单
/*
* Operation China Wall
* Author: Zuiho
* Date: 2018-05
* Problem: 2022-icebound's bill
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
using ll=long long;
using ull=unsigned long long;
int main()
{
ll n; cin>>n;
ll sum=0;
rep(i,1,n)
{
int tmp; cin>>tmp;
sum+=tmp;
}
if(sum>0) cout<<"icebound is happy.";
else if(sum==0) cout<<"icebound is ok.";
else cout<<"icebound is sad.";
return 0;
}ps：估计下一届的难度不会这么easy了2333
单机车七题还是难度不足的，不过也是考虑到广大兄弟院校了。
如非注明，原创内容遵循GFDLv1.3发布；其中的代码遵循GPLv3发布。
