Submission #330108

#	Time	Username	Problem	Language	Result	Execution time	Memory
330108		ignaciocanta	Laugh Analysis (IOI16_laugh)	C++14	100 / 100	5 ms	1516 KiB

laugh

#include <bits/stdc++.h>
#include <laugh.h>
 
using namespace std;

 
using tint = long long;
using ld = long double;
 
#define forsn(i, s, n) for(int i = s; i < int(n); i++)
#define forn(i, n) forsn(i, 0, n)
#define trav(a, x) for(auto& a : x)
 
using vi = vector<int>;
using vl = vector<tint>;
using vb = vector<bool>;
#define pb push_back
#define pf push_front
#define rsz resize
#define all(x) begin(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sz(x) (int)(x).size()
#define ins insert
#define esta(x,c) ((c).find(x) != (c).end())
 
using pi = pair<int,int>;
using pl = pair<tint,tint>;
#define f first
#define s second
#define mp make_pair
 
#define DBG(x) cerr << #x << " = " << x << endl;
 
const int MOD = 1e9+7;
const tint mod = 998244353;
const int MX = 105; 
const tint INF = 1e18; 
const int inf = 2e9;
const ld PI = acos(ld(-1)); 
const ld eps = 1e-8;
 
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
 
template<class T> void remDup(vector<T> &v){ 
    sort(all(v)); v.erase(unique(all(v)),end(v));
}
 
template<class T> bool valid(T x, T y, T n, T m){
    return (0<=x && x<n && 0<=y && y<m);
}

tint cdiv(tint a, tint b) { return a/b+((a^b)>0&&a%b); } //redondea p arriba
int fdiv(int a, int b) { return a/b-((a^b)<0&&a%b); } //redonde p abajo
 
void NACHO(string name = "cbarn"){
    ios_base::sync_with_stdio(0); cin.tie(0);
    //freopen((name+".in").c_str(), "r", stdin);
    //freopen((name+".out").c_str(), "w", stdout);
}

int dp[100005][2];

int longest_laugh(std::string s)
{
    forn(i, sz(s)) dp[i][0] = dp[i][1] = -inf;
	if(s[0] == 'h') dp[0][0] = 1;
	if(s[0] == 'a') dp[0][1] = 1;
	forsn(i, 1, sz(s)){
		if(s[i] == 'h'){
			dp[i][0] = max(1, dp[i-1][1]+1);
		}else if(s[i] == 'a') dp[i][1] = max(1, dp[i-1][0]+1);
		else dp[i][0] = dp[i][1] = -inf;
	}	
	int maxi = 0;
	forn(i, sz(s)) maxi = max(maxi, max(dp[i][0], dp[i][1]));
	return maxi;
}

• Subtask 1: 21 / 21
• Subtask 2: 26 / 26
• Subtask 3: 53 / 53