KEMENTERIAN PENDIDIKAN DAN KEBUDAYAAN

Transkripsi

1 Hak Cipta Dilindungi Undang-undang SOLUSI UJIAN OLIMPIADE SAINS NASIONAL 2013 CALON PESERTA INTERNATIONAL OLYMPIAD IN INFORMATICS (IOI) 2014 HARI KE-1 INFORMATIKA Waktu : 5 jam KEMENTERIAN PENDIDIKAN DAN KEBUDAYAAN DIREKTORAT JENDERAL PENDIDIKAN MENENGAH DIREKTORAT PEMBINAAN SEKOLAH MENENGAH ATAS TAHUN 2013

2 Tebak Himpunan Batas Waktu: 1 detik Batas Memory: 32 MB #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <queue> #include <stack> #include <vector> #include <utility> #include <map> #define REP(a,b) for (int a=0; a<b; a++) #define FOR(a,b,c) for (int a=b; a<=c; a++) #define FORD(a,b,c) for (int a=b; a>=c; a--) #define RESET(a,b) memset(a, b, sizeof a) #define EPS 1e-9 #define INF #define LL long long #define F first #define S second #define MP make_pair #define PB push_back #define PII pair<int,int> #define PDD pair<double,double> #define puts("") using namespace std; int n,k,q; char sc[1000]; void on() bool ada[105]; RESET(ada,0); REP(i,n) printf("1 %d\n", i+1); scanf("%s", sc); if (sc[0]!= 't') ada[i] = 1; int cnt = 0; REP(i,n) if (ada[i]) cnt++; printf("%d", cnt); REP(i,n) if (ada[i]) printf(" %d", i+1); Solusi Hari-1-1

3 //must be true void ologn() int ki,ka,tgh,ans; //ONE ki = 1; ka = n; while (ki <= ka) tgh = (ki + ka) >> 1; //ada: tgh? printf("1 %d\n", tgh); int h = 0; scanf("%s", sc); if (sc[0]!= 't') h = 1; if (h) ans = tgh; ki = tgh + 1; else ka = tgh - 1; printf("%d", ans); FOR(i,1,ans) printf(" %d", i); //must be true void grim() //eleminate all 0s bool ada[105]; FOR(i,1,n) ada[i] = 1; FOR(i,1,n) FOR(j,i+1,n) printf("2 %d %d\n", i, j); scanf("%s", sc); if (sc[0] == 't') //haha! ada[i] = 0; ada[j] = 0; Solusi Hari-1-2

4 //odd 0 or even 0? int cnt = 0; int h0 = 0; FOR(i,1,n) if (ada[i]) cnt++; else h0++; printf("%d", cnt); FOR(i,1,n) if (ada[i]) printf(" %d", i); scanf("%s", sc); if (sc[0] == 'y') //sweet, even 0 return; else //some work needed, it is odd 0! int liar = -1; FOR(i,1,n) if (ada[i]) printf("%d", cnt-1); FOR(j,1,n) if (j!= i && ada[j]) printf(" %d", j); bool h = 0; scanf("%s", sc); if (sc[0] == 'y') return; int main() gets(sc); scanf("%d %d %d", &n, &k, &q); if (k == 1) if (q > n) on(); else ologn(); else grim(); return 0; Solusi Hari-1-3

5 Berbaris Sebelum Masuk Batas Waktu: 1 detik Batas Memory: 32 MB #include<iostream> #include<cstring> #include<cstdio> #include<vector> #include<queue> #include<utility> #include<cassert> #include<algorithm> #define S 1000 #define N 1000 using namespace std; char asal[100]; int kasus,banyak; pair<int,int> batas[n+5]; vector<int> orang[s+5]; bool ada[s+5]; int jawab[n+5],prev[n+5]; bool cf(const int &a,const int &b) return (batas[a] <= batas[b]); int main() assert(scanf("%s %d",asal,&kasus) == 2); assert(scanf("%d",&banyak) == 1); assert(1 <= banyak && banyak <= N); assert(0 <= kasus && kasus <= 4); assert(strcmp(asal,"subsoal") == 0); if (kasus == 3) assert(banyak <= 8); memset(ada,0,sizeof(ada)); for (int i=0,sem;i<banyak;++i) assert(scanf("%d %d %d",&sem,&batas[i].first,&batas[i].second) == 3); assert(0 <= batas[i].first && batas[i].first <= batas[i].second && batas[i].second <= banyak); assert(1 <= sem && sem <= S); assert(!ada[sem]); ada[sem] = true; orang[sem].push_back(i); for (int i=1;i<=s;++i) sort(orang[i].begin(),orang[i].end(),cf); int masuk = 0; for (int i=1;i<=s;++i) if (orang[i].size() == 0) continue; priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > > pq; Solusi Hari-1-4

6 for (int j=0;j<masuk;++j) prev[j] = jawab[j]; jawab[j] = 0; int indeks1 = 0; int indeks2 = 0; for (int j=0;j<masuk + orang[i].size();++j) while (indeks2 < (int)orang[i].size() && batas[orang[i][indeks2]].first == j) pq.push(make_pair(batas[orang[i][indeks2]].second,orang[i][indeks2])); ++indeks2; if (pq.size() > 0) jawab[j] = pq.top().second; pq.pop(); else jawab[j] = prev[indeks1]; ++indeks1; masuk += orang[i].size(); assert(masuk == banyak); printf("%d",jawab[0]+1); for (int i=1;i<banyak;++i) printf(" %d",jawab[i]+1); return 0; Solusi Hari-1-5

7 Lipat Kertas Batas Waktu: 0,2 detik Batas Memory: 32 MB #include <cstdio> #include <cstdlib> #include <stack> #include <cassert> #include <algorithm> using namespace std; int n,i,x,size,sub; int pos[ ],bracket[ ]; bool valid,udah[ ]; char s[100]; int st[ ]; int main() scanf("%s%d",s,&sub); // Kasus scanf("%d",&n); if (sub == 3) assert(2 <= n && n <= 1000); if (sub == 4) assert(2 <= n && n <= 4); if (sub == 5) assert(2 <= n && n <= 1000); if (sub == 6) assert(2 <= n && n <= ); for (i=1 ; i<=n ; i++) scanf("%d",&x); assert(1 <= x && x <= n); //printf("%d\n",i); assert(udah[x] == false); udah[x] = true; pos[x] = i; // Simpan posisinya saja valid = true; for (i=2 ; i<=n ; i+=2) bracket[pos[i]] = i; bracket[pos[i-1]] = i; for (i=1 ; i<=n ; i++) if (bracket[i]!= 0) if (size == 0 bracket[i]!= st[size-1]) st[size] = bracket[i]; size++; else size--; if (size!= 0) valid = false; for (i=1 ; i<=n ; i++) bracket[i] = 0; size = 0; for (i=3 ; i<=n ; i+=2) bracket[pos[i]] = i; Solusi Hari-1-6

8 bracket[pos[i-1]] = i; for (i=1 ; i<=n ; i++) if (bracket[i]!= 0) if (size == 0 bracket[i]!= st[size-1]) st[size] = bracket[i]; size++; else size--; if (size!= 0) valid = false; if (!valid) printf("invalid\n"); else for (i=1 ; i<n ; i++) if (i % 2 == 1) if (pos[i] < pos[i+1]) printf("a"); else printf("b"); else if (pos[i] < pos[i+1]) printf("b"); else printf("a"); //sort(pos+1,pos+1+n); //for (i=1 ; i<=n ; i++) swap(pos[i],pos[(13*i+42)%100]); Solusi Hari-1-7

9 Menggelindingkan Kubus Batas Waktu: 0,2 detik Batas Memory: 32 MB #include <cstdio> #include <iostream> #include <vector> #include <algorithm> #include <stack> #include <queue> #include <map> #include <set> #include <utility> #include <sstream> #include <string> #include <iomanip> #include <cmath> #include <cstdlib> #include <cstring> using namespace std; #define fi first #define sc second #define MP make_pair #define pb push_back #define PI acos(-1.0) //alternative #define PI (2.0 * acos(0.0)) #define vi vector<int> #define vii vector<ii> #define ALL(c) (c).begin(), (c).end() #define RESET( c,a ) memset( (c), a, sizeof(c) ) #define REP( a,b,c ) for ( int a=b, _c=c; a<_c; ++a ) #define RED( a,b,c ) for ( int a=b, _c=c; a>=_c; --a ) #define REPI( it, c ) for ( typeof( (c).begin() ) it=(c).begin(); it!=(c).end(); ++it ) const int big = ; const double INF = 1e9; const double EPS = 1e-9; typedef long long LL; typedef pair<int,int> pii; typedef pair<ll,ll> pll; #define _DEBUG 1 #ifdef _DEBUG #define DEBUG printf #else #define DEBUG if (0) printf #endif // NTU The Lyons' Template // class dice public: int i[6]; Solusi Hari-1-8

10 dice() REP(x,0,6) i[x]=0; dice( int a[] ) REP(x,0,6) i[x]=a[x]; dice rot( int a ) int z = i[0]; dice ret; REP(x,0,6) ret.i[x] = i[x]; if (a==1) //utara ret.i[0] = i[4]; ret.i[4] = i[1]; ret.i[1] = i[2]; ret.i[2] = z; else if (a==2) //timur ret.i[0] = i[3]; ret.i[3] = i[1]; ret.i[1] = i[5]; ret.i[5] = z; else if (a==3) //selatan ret.i[0] = i[2]; ret.i[2] = i[1]; ret.i[1] = i[4]; ret.i[4] = z; else if (a==4) //barat ret.i[0] = i[5]; ret.i[5] = i[1]; ret.i[1] = i[3]; ret.i[3] = z; return ret; ; bool operator== (dice cmp) REP(x,0,6) if (i[x]!= cmp.i[x]) return false; return true; bool ud[8][8][8][8][8][8]; bool cek( dice a ) bool &lho = ud[a.i[0]][a.i[1]][a.i[2]][a.i[3]][a.i[4]][a.i[5]], ret = lho; if (!lho) lho = true; return ret; Solusi Hari-1-9

11 int main() int S; scanf("subsoal %d", &S); int i[6], h[6]; REP(x,0,6) scanf("%d", &h[x]); REP(x,0,6) scanf("%d", &i[x]); if (S==1) printf("2\n"); else if (S==2) printf("3\n"); else if (S==3) if ((i[0]==2) && (i[1]==2)) printf("0\n"); else printf("1\n"); else if (S==4) if (i[0]==2) printf("0\n"); else if (i[1]==2) printf("2\n"); else printf("1\n"); else RESET( ud,0 ); dice cur = dice(h), fin = dice(i); queue <dice> q; int o = 0; q.push( cur ); while (!q.empty()) int SZ = q.size(); while (SZ--) cur = q.front(); q.pop(); if (!cek(cur)) if (cur == fin) printf("%d\n",o); return 0; q.push( cur.rot(1) ); q.push( cur.rot(2) ); q.push( cur.rot(3) ); q.push( cur.rot(4) ); ++o; return 0; Solusi Hari-1-10