Submission #761155

#	Time	Username	Problem	Language	Result	Execution time	Memory
761155		fatemetmhr	Jousting tournament (IOI12_tournament)	C++17	0 / 100	1030 ms	736 KiB

tournament

//  ~ Be Name Khoda ~  //

#include <bits/stdc++.h>
//#pragma GCC optimize ("O3")
//#pragma GCC target("avx2")
//#pragma GCC optimize("unroll-loops,Ofast")

using namespace std;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second

const int maxn  =  1e6   + 10;
const int maxn5 =  5e5   + 10;
const int maxnt =  1.2e6 + 10;
const int maxn3 =  1e3   + 10;
const int mod   =  1e9   + 7;
const ll  inf   =  1e18;


int GetBestPosition(int n, int c, int r, int *k, int *s, int *e){

    for(int i = 0; i < n - 1; i++)
        k[i] = (k[i] > r);
    int pre = -1, nxt = 0;
    while(nxt < n - 1 && !k[nxt])
        nxt++;
    int mx = 0, out = 0;
    for(int i = 0; i < n; i++){
        if(i > nxt){
            pre = nxt;
            while(nxt < n - 1 && !k[nxt])
                nxt++;
        }
        int keepnxt = nxt, keeppre = pre;
        int l = max(0, pre), r = max(0, n - nxt - 2), pos = i, have = 0;
        nxt++;
        for(int j = 0; j < c; j++){
            //cout << "in " << j << ' ' << s[j] << ' ' << e[j] << ' ' << l << ' ' << r << ' ' << pos << ' ' << pre << ' ' << nxt << endl;
            if(e[j] <= pre){
                l -= e[j] - s[j];
                pre -= e[j] - s[j];
                nxt -= e[j] - s[j];
                pos -= e[j] - s[j];
            }
            else if(s[j] >= nxt)
                r -= e[j] - s[j];
            else if(s[j] <= pos && pos <= e[j]){
                if(s[j] <= pre || nxt <= e[j])
                    break;
                have++;
                nxt -= e[j] - s[j];
                pos = s[j];
            }
            else if(e[j] < pos){
                int vall = max(0, pre - s[j]);
                l -= vall;
                pre -= vall;
                pos -= e[j] - s[j];
                nxt -= e[j] - s[j];
            }
            else{
                int valr = max(0, e[j] - nxt);
                r -= valr;
                nxt -= e[j] - s[j] - valr;
            }
        }
        pre = keeppre;
        nxt = keepnxt;
        if(have > mx){
            mx = have;
            out = i;
        }
        //cout << i << ' ' << have << endl;
    }
    return out;

}

• Subtask 1: 0 / 17
• Subtask 2: 0 / 32
• Subtask 3: 0 / 51