제출 #518555

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
518555		sudheerays123	걷기 (NOI12_walking)	C++98	25 / 25	1 ms	336 KiB

walking

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define ll long long int
#define tc ll test;cin >> test;while(test--)
#define vi vector<ll>
#define pll pair<ll,ll>
#define pb push_back
#define mp make_pair
#define INF 1e18
#define MOD 1000000007
#define ff first
#define ss second
#define in >>
#define out <<
#define space << " " <<
#define spacef << " "
#define fo(i,a,b) for(ll i = a; i <= b; i++)
#define nextline out "\n"
#define print(x) for(auto i : x ) cout out i spacef
#define mmax(x,i) x = max(x,i)
#define mmin(x,i) x = min(x,i)
#define N 105

int main() {
    
    fast;

	ll l,n;
	cin in l in n;

	vector<pll> person(n+5);

	fo(i,1,n) cin in person[i].first in person[i].second;
	sort(person.begin()+1,person.begin()+n+1);

	vector<double> finish(n+5);
	fo(i,1,n) finish[i] = (double)l/(double)person[i].second + (double) person[i].first;

	vi dp(n+5); // dp[i] = max number of nested segments containing i

	// base case :

	fo(i,1,n) dp[i] = 1;

	// recursive relation :

	fo(i,1,n){
		fo(j,1,i-1){
			if(finish[j] > finish[i]) mmax(dp[i],dp[j]+1);
		}
	}

	// optimal solution :

	ll ans = -INF;
	fo(i,1,n) mmax(ans,dp[i]);

	cout out ans;
	
    return 0;
}

• Subtask 1: 3 / 3
• Subtask 2: 3 / 3
• Subtask 3: 5 / 5
• Subtask 4: 7 / 7
• Subtask 5: 7 / 7