Submission #591963

#	Time	Username	Problem	Language	Result	Execution time	Memory
591963		zaneyu	Meandian (CEOI06_meandian)	C++14	0 / 100	1 ms	320 KiB

meandian

/*input
6
2 9 9 6 6
7 1 9 10
9 2 5
4 10
8

*/
#include "libmean.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
//#pragma GCC target("avx2")
//order_of_key #of elements less than x
// find_by_order kth element
#define ll long long
#define ld long double
#define pii pair<int,int>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll maxn=100+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?" ":"\n");
    return out;
}
ll mult(ll a,ll b){
    return a*b%MOD;
}
ll mypow(ll a,ll b){
    a%=MOD;
    if(a==0) return 0;
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
int ans[maxn];
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n=Init();
    REP(i,n) ans[i]=-1;
    REP(asd,n-4){
        vector<int> v;
        REP(i,n) if(ans[i]==-1){
            v.pb(i);
            if(sz(v)==5) break;
        }
        vector<pii> vv;
        REP(i,5){
            vector<int> q;
            REP(j,5) if(i!=j) q.pb(j);
            vv.pb({Meandian(q[0],q[1],q[2],q[3]),v[i]});
        }
        sort(ALL(vv));
        ans[vv[2].s]=vv[0].f+vv[4].f-vv[2].f;
    }
    Solution(ans);
}

• Subtask 1: 0 / 100