Submission #1084325

#	Time	Username	Problem	Language	Result	Execution time	Memory
1084325		guymmk	Memory (IOI10_memory)	C++14	100 / 100	1 ms	596 KiB

memory

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include "grader.h"
#include "memory.h"
using namespace std;
using namespace __gnu_pbds;
//#pragma GCC optimize("O3,Ofast,unroll-loops")
//#pragma GCC target("popcnt")
#define int long long
#define ll long long
#define endl '\n'
#define pii pair<int,int>
#define debug(...) printf(__VA_ARGS__)
//#define debug(...)
#define trace(x) cout<<#x<<"="<<x<<endl;
//#define trace(x)
#define vi vector<int>
#define vpii vector<pii >
#define vvi vector<vector<int>
#define umap unordered_map
#define uset unordered_set
#define mk make_pair
#define pb push_back
#define pob pop_back
#define _ <<' '<<
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define read(x) for (auto& zu: (x)) cin >> zu;

static inline void print(const int &x) {
    if (x > 9)print(x / 10);
    putchar('0' + x % 10);
}

#define F first
#define S second
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define mustawa ios::sync_with_stdio(false);cin.tie(nullptr);

#define mishari main
const int inf=1e9+7;
void fa(bool ok){ cout << (ok ? "YES": "NO") << endl;}
int pw(ll a, ll b, ll mod= 1e9 + 7){
    ll result = 1;
    while(b) {
        if(b&1) {
            result = (result*a)%mod;
        }
        a = (a * a) % mod;
        b>>=1;
    }
    return result;
}
inline void usaco(string s) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
}int gcd(int a,int b){
    if (b == 0)return a;
    return gcd(b, a % b);
}

int lcm(int a, int b){
    return (a / gcd(a, b)) * b;
}
//long double log(int base,int x){
//    return log2(x)/log(base);
//}
int log(int base,int x){
    return ceil(log2(x)/log(base));
}
void vod(){}
void play() {
    vi p[26];
    for(int i=1;i<=50;i++) {
        char x = faceup(i);
        p[x - 'A'].pb(i);
    }for(int i=0;i<25;i++){
        faceup(p[i][0]);
        faceup(p[i][1]);
    }
}

• Subtask 1: 50 / 50
• Subtask 2: 50 / 50