hdu6852
题目
就是对于有最多k个逆序的反过来往后排
前面是正序的找到一个（正序的长度+（（n-正序的长度）%逆序的长度！=0）+（n-正序的长度）/逆序的长度））

#include <iostream>using namespace std;int main()
{ios::sync_with_stdio(0);cin.tie(0),cout.tie(0);int T;cin>>T;int flag;while(T--){flag=0;int n,x,y;cin>>n>>x>>y;int t=0;for(;t<=n;t++){if(t+((n-t)%y!=0)+(n-t)/y==x){break;}}if(t>n||(n-t)/y==0){cout<<"NO"<<"\n";continue;}cout<<"YES"<<"\n";for(int i=1;i<=t;i++){cout<<i<<" ";flag++;}if((n-t)%y!=0){for(int i=t+(n-t)%y;i>t;i--){cout<<i<<" ";flag++;}}int ss=(n-t)/y;while(ss--){for(int i=n-ss*y;i>=n+1-(ss+1)*y;i--){cout<<i;flag++;if(flag!=n)cout<<" ";}}cout<<"\n";}return 0;
}

hdu6853
题目

#include <bits/stdc++.h>using namespace std;
typedef long long ll;int dir[8][2]={0,1,0,-1,1,0,-1,0,1,1,1,-1,-1,1,-1,-1};
ll x,y;
int flag;
ll ans,bns;
map<pair<ll,ll>,int>M;
queue <pair<ll,ll> > Q;void bfs()
{pair<ll,ll> head,next;head.first=x;head.second=y;Q.push(head);M[head]=1;while(!Q.empty()){head=Q.front();if(head.first==head.second){cout<<"0/1"<<endl;return;}ans++;Q.pop();for(int i=0;i<8;i++){next.first=head.first+dir[i][0];next.second=head.second+dir[i][1];if(__gcd(next.first,next.second)>1){ans++;if(M[next]==0){Q.push(next);M[next]=1;}}}if(flag==0){bns=ans;flag=1;}}ll g=__gcd(ans,bns);cout<<bns/g<<"/"<<ans/g<<"\n";
}
int main()
{ios::sync_with_stdio(0);int T;cin>>T;while(T--){M.clear();flag=0;bns=0;ans=0;cin>>x>>y;while(!Q.empty()){Q.pop();}bfs();}return 0;
}

hdu6850
题目

#include <bits/stdc++.h>
using namespace std;
const int maxn=2005;
typedef long long ll;
struct node
{ll x,y;
};
node no[maxn];
int visit[maxn];
set<int>vv;
struct edge
{int a,b;ll dis;bool operator<(const edge & bb)const{return dis>bb.dis;}
};
edge e[maxn*maxn];
void init()
{memset(visit,0,sizeof(visit));vv.clear();
}
ll getdis(int i,int j)
{return (no[i].x-no[j].x)*(no[i].x-no[j].x)+(no[i].y-no[j].y)*(no[i].y-no[j].y);
}
void solve()
{int n;cin>>n;for(int i=1;i<=n;i++){ll x,y;cin>>x>>y;no[i].x=x;no[i].y=y;}int res=0;for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){res++;e[res].a=i;e[res].b=j;e[res].dis=getdis(i,j);}}sort(e+1,e+1+res);ll max_=e[1].dis;for(int i=1;i<=res;i++){if(visit[e[i].a]||visit[e[i].b])continue;if(e[i].dis==max_){int a=e[i].a;int b=e[i].b;vv.insert(a);vv.insert(b);} else{if(vv.find(e[i].a)!=vv.end()||vv.find(e[i].b)!=vv.end())continue;for(auto m:vv)visit[m]=1;vv.clear();vv.insert(e[i].a);vv.insert(e[i].b);max_=e[i].dis;}}for(auto m:vv)visit[m]=1;int flag=0;for(int i=1;i<=n;i++){if(visit[i]==0){flag=i;break;}}if(flag==1)cout<<"NO"<<endl;elsecout<<"YES"<<endl;}
int main() {int _=1;cin>>_;while(_--){init();solve();}return 0;
}