#ctpc

[itsazodiacthing]

CANCER PARENT & TAURUS CHILD

This is usually a good combination. Both signs tend to have a strong emotional bond and a good understanding of each other. These two signs have a lot in common, and both long for security and warmth in relationships. Usually, they form a very tightly knit family. They both appreciate a solid home base and the material comforts of domestic life. Cancer, a water sign, is a born nurturer, and takes excellent care of the Taurus child. Both parent and child enjoy being at home, love each other`s company. But problems could arise if the Taurus child becomes stubborn causing the Cancer parent to sulk. As a result, this child needs to understand the parent`s emotional sensitivity, and the parent needs to be more open about his/her needs instead of relying on emotional blackmail. The Taurus-Cancer relationship is a solid, steady and reliable union. Both are dependable and strongly oriented toward family life. A mutual love of security and the home, makes theirs an ideal family relationship.

[yt-siden-algo]

CTPC:K Prime Factorial

解法

まず，Erathosthenesの篩いなどによって\( 10^7 \)未満の素数を求めておく． あとは，小さい順に掛けていき，判定するだけ．

コード

#include <iostream> #include <vector> using namespace std; const int MAX = 10000000; const int MOD = 1000000; bool p[MAX+1]; int main() { int n; cin >> n; vector<int> pr; for (int i=0; i<=MAX; i++) p[i]=1; p[0] = p[1] = 0; for (int i=2; i*i<=MAX; i++) { if (p[i]) { for (int k=i+i; k<=MAX; k+=i) p[k] = 0; } } for (int i=2; i<=MAX; i++) if (p[i]) pr.push_back(i); for (int i=0; i<n; i++) { long long num=1; int c; cin >> c; bool f=0; for (int k=0; !f&&k<pr.size(); k++) { num *= pr[k]; num %= MOD; if (num == c) { f=1; break; } } cout << (f ? "Yes" : "No") << endl; } }

[yt-siden-algo]

CTPC:G Lucky Number

解法

ビットマスクを利用する．すなわち，\( n \)番目の数は， \[ a(i) = 7^i \hspace{1em} (for \hspace{1ex} i \ge 0),\hspace{1em} N = \lceil \log_2n \rceil \hspace{1em} (for \hspace{1ex} n \ge 1) \] とすると， \[ \sum_{i=0}^{N-1} {f(i,n)}, \hspace{1em} where \hspace{1ex} f(i,n) = \left\{ \begin{array}{cc} a(i) & n \& (1 \ll i) \neq 0 \\\\ 0 & otherwise \end{array} \right. \] で表される．(\( \& \)はビット論理積，\( \ll \)はビット左シフトとします) あと，求めるものは\( 10^6 \)のmodなので\( a(i) \)もmodで求めておく．

コード

#include <iostream> using namespace std; typedef long long ll; const int MOD = 1000000; int main() { ll n; cin >> n; ll N=1; while ((1<<(N-1)) < n) { N++; } ll a[N]; a[0] = 1; for (ll i=1; i<N; i++) { a[i] = a[i-1] * 7; a[i] %= MOD; } ll ans = 0; for (int i=0; i<N; i++) if (n&(1<<i)) ans = (ans + a[i]) % MOD; cout << ans << endl; }

[yt-siden-algo]

CTPC:E Airport

解法

まず入力された情報をもとに隣接行列を作り，全ての互いに違う空港のペアA,BについてA→BまたはB→Aとなっているかチェックする． 次に，Floyd-Warshallで全点間が接続されていることを確認する．(普通にTSPでも大丈夫らしい)

コード

#include <iostream> #include <map> #include <string> #include <cstring> using namespace std; const int MAX = 1<<20; int main() { bool tab[20][20]; memset(tab, 0, sizeof(tab)); int n; cin >> n; string s; map<string, int> dic; for (int i=0; i<n; i++) { cin >> s; dic[s] = i; } int S; cin >> S; for (int i=0; i<S; i++) { string s1, s2; cin >> s1 >> s2; tab[ dic[s1] ][ dic[s2] ] = 1; } bool f=0; for (int i=0; i<n-1; i++) for (int k=i+1; k<n; k++) if (!(tab[i][k]^tab[k][i])) { f=1; break; } if (f) { cout << "No" << endl; return 0; } int fw[n][n]; for (int i=0; i<n; i++) for (int k=0; k<n; k++) if (i==k) fw[i][i] = 0; else if (tab[i][k]) fw[i][k] = 1; else fw[i][k] = MAX; for (int w=0; w<n; w++) for (int i=0; i<n; i++) for (int k=0; k<n; k++) fw[i][k] = min(fw[i][k], fw[i][w]+fw[w][k]); f=0; for (int i=0; !f&&i<n; i++) { for (int k=0; !f&&k<n; k++) { if (i==k) continue; if (fw[i][k] == MAX) { f=1; break; } } } cout << (f ? "No" : "Yes") << endl; }

[yt-siden-algo]

CTPC:B Macaron

解法

実装するだけ．反転する点がスタンプの中心からのチェビシェフ距離であることに気が付けば非常に簡単に書くことができる．(あとで教えてもらった)

コード

#include <iostream> #include <cstring> using namespace std; int dx[] = { 1, 0, -1, 0 }; int dy[] = { 0, 1, 0, -1 }; int main() { int H, W, K; cin >> W >> H >> K; int X[K], Y[K], L[K]; for (int i=0; i<K; i++) cin >> X[i] >> Y[i] >> L[i]; bool tab[H][W]; memset(tab, 0, sizeof(tab)); for (int i=0; i<K; i++) { bool f=1; for (int n=1; f; n++) { f=0; int siz = 2*n*(L[i]+1)-1; for (int x=X[i]-siz/2; x<=X[i]+siz/2; x++) { int y1=Y[i]-siz/2, y2=Y[i]+siz/2; if (0<=x&&x<W&&0<=y1&&y1<H) { tab[y1][x] ^= true; f=1; } if (0<=x&&x<W&&0<=y2&&y2<H) { tab[y2][x] ^= true; f=1; } } for (int y=Y[i]-siz/2+1; y<Y[i]+siz/2; y++) { int x1=X[i]-siz/2, x2=X[i]+siz/2; if (0<=x1&&x1<W&&0<=y&&y<H) { tab[y][x1] ^= true; f=1; } if (0<=x2&&x2<W&&0<=y&&y<H) { tab[y][x2] ^= true; f=1; } } } } for (int y=0; y<H; y++) { for (int x=0; x<W; x++) cout << (tab[y][x] ? '#' : '.'); cout << endl; } }

[yt-siden-algo]

CTPC:A Average

解法

やるだけ．

コード

#include <iostream> using namespace std; int main() { int s = 0; int m = 0; int t; for (int i=0; i<4; i++) { cin >> t; s += t; m = max(m, t); } s += m; s /= 5; cout << (s>=60 ? "Yes" : "No") << endl; }