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SDUT 2021 Winter Individual Contest

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SDUT 2021 Winter Individual Contest

Gym-102448

  • A - Accept or Reject
  • B - Beza's Hangover
  • D - Drinking to turn red
  • E - Everybody loves acai
  • F - Finally, christmas!
  • G - Gorggeous Peter's Great Friend
  • K - Kongey Donk
  • H - Hellcife is on fire(补)
  • I - Ivan and the swimming pool(补)

A - Accept or Reject

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 1e6 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;int n,m;
string s;
bool flag;
int dp[N];
char str[N];int get(string s){int len=s.size();str[0]='$';str[1]='#';int pos=2;rep(i,0,len-1){str[pos++]=s[i];str[pos++]='#';}str[pos]=0;return pos;
}void manacher(string s){int len=get(s);int maxn=0;int pos=0;rep(i,1,len-1){if(i<maxn) dp[i]=min(maxn-i,dp[2*pos-i]);else dp[i]=1;while(str[i+dp[i]]==str[i-dp[i]]){dp[i]++;if(maxn<i+dp[i]){pos=i;maxn=i+dp[i];}}}
}void solve(){//cout<<(N<<1)<<endl;cin>>n>>m;cin>>s;manacher(s);flag=0;int len=2*(int)s.size();rep(i,0,len-1){if(m+1<=dp[i]&&(m+1)%2==dp[i]%2) flag=1;}if(flag) puts("Accept");else puts("Reject");
}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

B - Beza’s Hangover

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 2e5 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;int n,m,q;
string s[N];
string str;
int vis[N];
map<string,int>mp;void get(int k,int sum){for(;k<=n;k+=(lowbit(k))) vis[k]+=sum;
}int query(int k){if(!k) return 0;int res=0;for(;k;k-=(lowbit(k))) res+=vis[k];return res;
}void solve(){cin>>n>>m>>q;rep(i,1,n) cin>>s[i];rep(i,1,m){int k;cin>>str>>k;mp[str]=k;}rep(i,1,n) get(i,mp[s[i]]);while(q--){int k;cin>>k;if(k==1){int p;cin>>p>>str;int sum=query(p)-query(p-1);get(p,mp[str]-sum);}else{int l,r;cin>>l>>r;if(query(r)-query(l-1)>30*(r-l+1)) puts("YES");else puts("NO");}}
}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

D - Drinking to turn red

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 2e5 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;struct node{double d;int r;
}dp[N];bool cmp(node x,node y){return x.d<y.d;
}int n,x,y;void solve(){cin>>n>>x>>y;rep(i,1,n){int a,b,r;cin>>a>>b>>r;ll d=(ll)(x-a)*(x-a)+(ll)(y-b)*(y-b);dp[i].d=sqrt(d)-r;dp[i].r=r;}sort(dp+1,dp+1+n,cmp);double st=0.0;double res=0.0;rep(i,1,n){if(st<dp[i].d){res+=dp[i].d-st;st+=dp[i].d-st;}st+=dp[i].r;}printf("%.10f\n",res);
}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

E - Everybody loves acai

题目链接

不用scanf我是T了 （T^T）

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 1e6 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;int dp[]={6,28,496,8128};void solve(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int ans;if(n<dp[0]) ans=-1;else if(n<dp[1]) ans=dp[0];else if(n<dp[2]) ans=dp[1];else if(n<dp[3]) ans=dp[2];else ans=dp[3];printf("%d\n",ans);}
}
int main() {solve();return 0;
}

F - Finally, christmas!

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 2e5 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;struct node{ll h;ll x;ll num;
}dp[N];bool cmp(node x,node y){if(x.x<y.x) return 1;else if(x.x==y.x&&x.num<y.num) return 1;return 0;
}int pos;
multiset<ll>mp;void add(ll h,ll x,ll num){dp[pos].h=h;dp[pos].x=x;dp[pos++].num=num;
}void solve(){ll n;cin>>n;rep(i,1,n){int l,r,h;cin>>l>>r>>h;add(h,l,0);add(h,r,1);}sort(dp,dp+pos,cmp);mp.insert(dp[0].h);ll res=0;rep(i,1,pos-1){if(mp.size()){ll key=*(--mp.end());res+=key*(dp[i].x-dp[i-1].x);}if(!dp[i].num) mp.insert(dp[i].h);else mp.erase(mp.find(dp[i].h));}cout<<res<<endl;
}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

G - Gorggeous Peter’s Great Friend

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 1e6 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;int c,p,s;
string str;
vector<string>vp;
map<string,int> dp;
map<string,int> mp;void solve(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>c>>p>>s;rep(i,0,c-1){cin>>str;vp.pb(str);dp[str]=0;}rep(i,0,p-1){int k;cin>>str>>k;mp[str]=k;}string a,b;rep(i,0,s-1){cin>>str>>a>>b;if(b=="AC") dp[str]+=mp[a];}vector<string>::iterator it;for(it=vp.begin();it!=vp.end();it++){cout<<*it<<" "<<dp[*it]<<endl;}
}
int main() {solve();return 0;
}

K - Kongey Donk

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 2e5 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;int n,h;
vector<ll> dp[N];
vector<ll> mp[N];void init(){rep(i,1,n){rep(j,1,h){mp[i].pb(0);}}
}void solve(){cin>>n>>h;int k;rep(i,1,n){rep(j,1,h){cin>>k;dp[i].pb(k);}}init();rep(i,1,n) mp[i][0]=dp[i][0];rep(j,1,h-1){rep(i,1,n){if(i>1) mp[i][j]=max(mp[i][j],mp[i-1][j-1]+dp[i][j]);if(i<n) mp[i][j]=max(mp[i][j],mp[i+1][j-1]+dp[i][j]);mp[i][j]=max(mp[i][j],mp[i][j-1]+dp[i][j]);}}ll maxn=-1;int pos=h-1;rep(i,1,n) maxn=max(maxn,mp[i][pos]);cout<<maxn<<endl;
}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

H - Hellcife is on fire(补)

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 2e5 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
using namespace std;struct node{ll w;ll to;ll next;
}dp[N];struct NODE{ll t;ll x;friend bool operator <(const NODE& a,const NODE& b){return a.t>b.t;}
};ll n,m,k;
ll num;
ll Next[N];
ll st[N];
ll vp[N];
priority_queue<NODE>que;void init(){num=0;memset(Next,-1,sizeof(Next));
}void add(ll u,ll v,ll w){dp[num].to=v;dp[num].w=w;dp[num].next=Next[u];Next[u]=num++;
}void solve(){init();cin>>n>>m>>k;rep(i,1,m){ll u,v,w;cin>>u>>v>>w;add(u,v,w);add(v,u,w);}rep(i,1,n){vp[i]=1e18;cin>>st[i];}rep(i,1,k){ll k;cin>>k;vp[k]=st[k];que.push(NODE{vp[k],k});}while(!que.empty()){ll topx=que.top().x;ll topt=que.top().t;que.pop();for(ll i=Next[topx];i!=-1;i=dp[i].next){ll to=dp[i].to;ll w=dp[i].w;if(vp[to]>topt+w+st[to]){vp[to]=topt+w+st[to];que.push(NODE{vp[to],to});}}}rep(i,1,n) cout<<vp[i]<<endl;
}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

I - Ivan and the swimming pool(补)

题目链接

答案：

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f3f
#define rep(i,a,b) for(auto i=a;i<=b;++i)
#define bep(i,a,b) for(auto i=a;i>=b;--i)
#define lowbit(x) x&(-x)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define PI acos(-1)
#define pb push_back
#define eps 1e-6
const int mod = 1e9 + 7;
const int MOD = 20210213;
const int N = 2e5 + 10;
const int M = 55;
int dx[]={-1, 0, 1, 0};
int dy[]={0, 1, 0, -1};
int dxy[][2]={{0,1},{0,-1},{1,0},{-1,0}};
using namespace std;vector<vector<int> >v1;
vector<vector<int> >v2;struct node{int x;int y;
};int BFS(int x,int y,int n,int m){int res=1;v2[x][y]=0;queue<node>q;q.push((node){x,y});while(!q.empty()){node p=q.front();q.pop();rep(i,0,3){int xx=p.x+dxy[i][0];int yy=p.y+dxy[i][1];if(xx>0&&yy>0&&xx<=n&&yy<=m&&v2[xx][yy]>0){v2[xx][yy]=0;res++;q.push((node){xx,yy});}}}return res;
}bool judge(int s,int n,int m,int d){rep(i,1,n){rep(j,1,m){if(v1[i][j]>=d) v2[i][j]=1;else v2[i][j]=0;}}int maxn=0;rep(i,1,n){rep(j,1,m){if(v2[i][j]) maxn=max(maxn,BFS(i,j,n,m));}}if(maxn>=s) return 1;else return 0;
}void solve(){int s,n,m;cin>>s>>n>>m;v1.resize(n+10);v2.resize(n+10);rep(i,0,n){v1[i].resize(m+10);v2[i].resize(m+10);}int l=mod;int r=0;rep(i,1,n){rep(j,1,m){cin>>v1[i][j];l=min(l,v1[i][j]);r=max(r,v1[i][j]);}}l--;r++;while(r-l>1){int mid=(l+r)>>1;if(judge(s,n,m,mid)) l=mid;else r=mid;}cout<<l<<endl;
}int main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);solve();return 0;
}

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SDUT 2021 Winter Individual Contest

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