kyoupuro @ ウィキ

When Can We Meet?

2013-01-22T17:24:19+09:00

**1124 : When Can We Meet? [[問題文>http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1124]] ***解説なるべく多くの人数が参加できるような日程を探す問題。何日に何人が参加できるかを記憶しておき、それを元に参加人数が最大でかつ最短の日にちを探索すればいい。 ***プログラム ***C ***C++ #region #include using namespace std; int main() { int n, q; while (cin >> n >> q) { if (n == 0 && q == 0) break; int day[100] = {0}; int m, d; for (int i = 0; i < n; ++i) { cin >> m; for (int j = 0; j < m; ++j) { cin >> d; ++day[d]; } } int max = 0, max_i = 0; for (int i = 0; i < 100; ++i) { if (max < day[i]) { max = day[i]; max_i = i; } } if (max >= q) { cout << max_i << endl; } else { cout << 0 << endl; } } return 0; } #endregion ***Java

Volume11

2013-01-22T17:19:22+09:00

**Volume11 |問題番号|問題名|C|C++|Java| |1100|[[Area of Polygons]]|-|○|-| |1101||-|-|-| |1102||-|-|-| |1103||-|-|-| |1104||-|-|-| |1105||-|-|-| |1106||-|-|-| |1107||-|-|-| |1108||-|-|-| |1109||-|-|-| |1110||-|-|-| |1111||-|-|-| |1112||-|-|-| |1113||-|-|-| |1114||-|-|-| |1115||-|-|-| |1116||-|-|-| |1117||-|-|-| |1118||-|-|-| |1119|[[Exploring Caves]]|-|○|-| |1120||-|-|-| |1121||-|-|-| |1122||-|-|-| |1123||-|-|-| |1124|[[When Can We Meet?]]|-|○|-| |1125||-|-|-| |1126||-|-|-| |1127||-|-|-| |1128||-|-|-| |1129||-|-|-| |1130|[[Red and Black]]|-|○|-| |1131||-|-|-| |1132||-|-|-| |1133||-|-|-| |1134||-|-|-| |1135||-|-|-| |1136||-|-|-| |1137||-|-|-| |1138||-|-|-| |1139||-|-|-| |1140||-|-|-| |1141|[[Dirichlet's Theorem on Arithmetic Progressions]]|-|○|-| |1142|[[Organize Your Train part II]]|-|○|-| |1143||-|-|-| |1144|[[Curling 2.0]]|-|○|-| |1145||-|-|-| |1146||-|-|-| |1147||-|-|-| |1148||-|-|-| |1149||-|-|-| |1150||-|-|-| |1151||-|-|-| |1152||-|-|-| |1153||-|-|-| |1154||-|-|-| |1155||-|-|-| |1156||-|-|-| |1157||-|-|-| |1158||-|-|-| |1159||-|-|-| |1160||-|-|-| |1161||-|-|-| |1162||-|-|-| |1163||-|-|-| |1164||-|-|-| |1165||-|-|-| |1166||-|-|-| |1167||-|-|-| |1168||-|-|-| |1169||-|-|-| |1170||-|-|-| |1171||-|-|-| |1172||-|-|-| |1173||-|-|-| |1174||-|-|-| |1175||-|-|-| |1176||-|-|-| |1177||-|-|-| |1178||-|-|-| |1179||-|-|-| |1180||-|-|-| |1181||-|-|-| |1182||-|-|-| |1183||-|-|-| |1184||-|-|-| |1185|Patisserie ACM|-|-|-|

Exploring Caves

2013-01-17T20:06:29+09:00

**1119 : Exploring Caves [[問題文>http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1119]] ***解説スタート位置から最も遠い部屋の座標を答える問題。今の位置からのどれだけ移動するかが入力で与えられるので、その通りに移動して逐一距離を計算すればいい。英語の勉強しましょう。 ***プログラム ***C ***C++ #region #include using namespace std; int main(){ int n; cin >> n; for (int i = 0; i < n; i++) { int x = 0, y = 0; int dx, dy; int max = 0, tx, ty; while (cin >> dx >> dy, dx || dy) { x += dx; y += dy; int d = x*x + y*y; if (max < d) { max = d; tx = x; ty = y; } else if (max == d && tx < x) { tx = x; ty = y; } } cout << tx << " " << ty << endl; } return 0; } #endregion ***Java

Flying Jenny

2013-01-14T16:28:27+09:00

**0082 : Flying Jenny [[問題文>http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0082]] ***解説条件に合うようなメリーゴーランドの並べ方を出力する。メリーゴーランドの並び方は8種類なので、先に並び方を全て配列に格納しておき、小さいほうから順に調べていけばいい。ただ、最小値を求めるとき、最初の値が小さいとWAになってしまうので注意([[10000]]でもWAになった)。 ***プログラム ***C ***C++ #region #include #include #include using namespace std; string pos[] = { "12121414", "12141412", "14121214", "14141212", "21214141", "21414121", "41212141", "41412121", }; int main() { int n[8]; while (cin >> n[0] >> n[1] >> n[2] >> n[3] >> n[4] >> n[5] >> n[6] >> n[7]) { int min = INT_MAX; string ans; for (int i = 0; i < 8; i++) { int sum = 0; for (int j = 0; j < 8; j++) { int a = n[j] - (pos[i][j] - '0'); if (a < 0) a = 0; sum += a; } if (min > sum) { ans = pos[i]; min = sum; } } cout << ans[0]; for (int i = 1; i < 8; i++) { cout << " " << ans[i]; } cout << endl; } return 0; } #endregion ***Java

Volume0

2013-01-14T16:22:15+09:00

|問題番号|問題名|C|C++|Java| |0000|[[QQ]]|-|○|-| |0001|[[List of Top 3 Hills]]|○|○|-| |0002|[[Digit Number]]|-|○|○| |0003|[[Is it a Right Triangle?]]|-|○|-| |0004|[[Simultaneous Equation]]|-|○|-| |0005|[[GCD and LCM]]|-|○|-| |0006|[[Reverse Sequence]]|-|○|-| |0007|[[Debt Hell]]|-|○|-| |0008|[[Sum of 4 Integers]]|-|○|-| |0009|[[Prime Number]]|-|○|-| |0010|[[Circumscribed Circle of a Triangle]]|-|○|-| |0011|[[Drawing Lots]]|-|○|-| |0012||-|-|-| |0013|[[Switching Railroad Cars]]|-|○|-| |0014||-|-|-| |0015||-|-|-| |0016||-|-|-| |0017||-|-|-| |0018||-|-|-| |0019||-|-|-| |0020||-|-|-| |0021||-|-|-| |0022||-|-|-| |0023||-|-|-| |0024||-|-|-| |0025||-|-|-| |0026||-|-|-| |0027||-|-|-| |0028||-|-|-| |0029||-|-|-| |0030||-|-|-| |0031||-|-|-| |0032||-|-|-| |0033||-|-|-| |0034||-|-|-| |0035|[[Is it Convex?]]|-|○|-| |0036||-|-|-| |0037||-|-|-| |0038||-|-|-| |0039||-|-|-| |0040||-|-|-| |0041||-|-|-| |0042||-|-|-| |0043|[[Puzzle]]|-|○|-| |0044||-|-|-| |0045||-|-|-| |0046||-|-|-| |0047||-|-|-| |0048||-|-|-| |0049||-|-|-| |0050||-|-|-| |0051||-|-|-| |0052||-|-|-| |0053||-|-|-| |0054||-|-|-| |0055||-|-|-| |0056||-|-|-| |0057||-|-|-| |0058||-|-|-| |0059||-|-|-| |0060||-|-|-| |0061||-|-|-| |0062||-|-|-| |0063||-|-|-| |0064||-|-|-| |0065||-|-|-| |0066||-|-|-| |0067||-|-|-| |0068||-|-|-| |0069||-|-|-| |0070||-|-|-| |0071||-|-|-| |0072||-|-|-| |0073||-|-|-| |0074||-|-|-| |0075||-|-|-| |0076||-|-|-| |0077||-|-|-| |0078||-|-|-| |0079||-|-|-| |0080||-|-|-| |0081||-|-|-| |0082|[[Flying Jenny]]|-|○|-| |0083||-|-|-| |0084||-|-|-| |0085|[[Joseph's Potato]]|-|○|-| |0086||-|-|-| |0087||-|-|-| |0088|[[The Code A Doctor Loved]]|-|○|-| |0089||-|-|-| |0090||-|-|-| |0091||-|-|-| |0092||-|-|-| |0093|[[Leap Year]]|-|○|-| |0094||-|-|-| |0095||-|-|-| |0096||-|-|-| |0097||-|-|-| |0098||-|-|-| |0099||-|-|-|

Joseph's Potato

2013-01-14T15:44:09+09:00

**0085 : Joseph's Potato [[問題文>http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0085]] ***解説 Joseph's Potatoというゲームのルールに従って、最後に残る人の番号を答える問題。一人になるまでループを使ってシミュレーションすればいい。どの番号が残っているかという情報を記憶する配列があると楽。 ***プログラム ***C ***C++ #region #include using namespace std; int main() { int m, n; while (cin >> m >> n, m || n) { int p[1000] = {0}, cnt = 1, res = m, num; for (int i = 0; ; i++) { if (res == 1 && p[i%m] == 0) { num = i%m; break; } if (p[i%m]) continue; if (cnt == n) { p[i%m] = 1; res--; cnt = 1; } else { cnt++; } } cout << num+1 << endl; } return 0; } #endregion ***Java

Emacs-like Editor

2013-01-13T16:44:04+09:00

**1021 : Emacs-like Editor [[問題文>http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1021]] ***解説コマンドによって何度か文字列の操作を行い、その結果を出力する。書いてある通りにコーディングすればいい・・・が、よく分からないところでWAになりやすい。とりあえず、・最初に与えられる文章を一行ずつ読み込む（空白が含まれている可能性がある）・コマンドYでバッファに改行があるときの処理に注意するあたりが気をつけるべきポイントか。最悪、プログラムを見比べながら間違いを見つけてください・・・ ***プログラム ***C ***C++ #region #include #include #include #include #include using namespace std; int main() { vector v; for (int i = 0; ; i++) { string w; getline(cin, w); if (w == "END_OF_TEXT") break; v.push_back(w); } string c, buf; int x = 0, y = 0; while (cin >> c) { if (c == "-") break; if (c == "a") { x = 0; } else if (c == "e") { x = v[y].size(); } else if (c == "p") { if (y >= 1) y--; x = 0; } else if (c == "n") { if (y < v.size()-1) y++; x = 0; } else if (c == "f") { if (x != v[y].size()) x++; else if (y < v.size()-1 && x == v[y].size()) { y++; x = 0; } } else if (c == "b") { if (x != 0) x--; else { if (y >= 1) { y--; x = v[y].size(); } } } else if (c == "d") { if (x != v[y].size()) { v[y] = v[y].substr(0, x) + v[y].substr(x+1); } else { if (y < v.size()-1) { v[y] += v[y+1]; for (int i = y+1; i < v.size()-1; i++) v[i] = v[i+1]; v.pop_back(); } } } else if (c == "k") { if (x != v[y].size()) { buf = v[y].substr(x); v[y] = v[y].substr(0, x); } else { if (y < v.size()-1) { v[y] += v[y+1]; for (int i = y+1; i < v.size()-1; i++) v[i] = v[i+1]; v.pop_back(); buf = "RETURN"; } } } else if (c == "y") { if (buf.empty()) { // 何もしない } else if (buf == "RETURN") { string w = ""; v.push_back(""); for (int i = v.size()-1; i > y+1; i--) { v[i] = v[i-1]; } if (x != v[y].size()) { w = v[y].substr(x); v[y] = v[y].substr(0, x); } y++; x = 0; } else { v[y] = v[y].substr(0, x) + buf + v[y].substr(x); x += buf.size(); } } } for (int i = 0; i < v.size(); i++) { cout << v[i] << endl; } return 0; } #endregion ***Java ・WAになったプログラム #region #include #include using namespace std; const int MAX = 100; int main(){ string s[MAX+1]; for (int i = 0; i <= MAX; i++) s[i] = ""; for (int i = 0; ; i++) { string w; cin >> w; if (w == "END_OF_TEXT") break; s[i] += w + "!"; } string c, buf = ""; int x = 0, y = 0; while (cin >> c) { if (c == "-") break; if (c == "a") { x = 0; } else if (c == "e") { x = s[y].size()-1; } else if (c == "p") { x = 0; if (y > 0) { y--; } } else if (c == "n") { x = 0; if (s[y+1] != "") { y++; } } else if (c == "f") { if (x == s[y].size()-1) { if (s[y+1] != "") { x = 0; y++; } } else { x++; } } else if (c == "b") { if (x == 0) { if (y > 0) { y--; x = s[y].size()-1; } } else { x--; } } else if (c == "d") { if (x == s[y].size()-1) { // 行末にある if (s[y+1] != "") { s[y] = s[y].substr(0, s[y].size()-1) + s[y+1]; for (int i = y+1; i <= MAX-1; i++) { s[i] = s[i+1]; } } } else { // 行末にない s[y] = s[y].substr(0, x) + s[y].substr(x+1); } } else if (c == "k") { if (x == s[y].size()-1) { // 行末にある if (s[y+1] != "") { s[y] += s[y+1]; for (int i = y+1; i <= MAX; i++) { s[i] = s[i+1]; } buf = "!"; } } else { // 行末にない buf = s[y].substr(x, s[y].size()-x-1); s[y] = s[y].substr(0, x) + "!"; x = s[y].size()-1; } } else if (c == "y") { if (buf != "") { if (buf == "!") { // 改行 string w = s[y].substr(x); s[y] = s[y].substr(0, x-1); for (int i = MAX; i >= y+2; i--) s[i] = s[i-1]; s[y+1] = w; y = y+1; x = 0; } else { s[y] = s[y].substr(0, x) + buf + s[y].substr(x); x += buf.size(); } } } cout << "x = " << x << endl; cout << "y = " << y << endl; cout << "buf = " << buf << endl; for (int i = 0; s[i] != ""; i++) { cout << s[i] << endl; } cout << endl; } for (int i = 0; s[i] != ""; i++) { cout << s[i].substr(0, s[i].size()-1) << endl; } return 0; } #endregion

Cleaning Robot

2013-01-08T16:11:10+09:00

**1020 : Cleaning Robot [[問題文>http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=1020]] ***解説ロボットが移動した後に指定された場所にいるか確率を答える問題。全探索しようとすると時間がかかりすぎるので、各部屋ごとにその部屋にいける場合の数をカウントする。 ***プログラム ***C ***C++ #region #include #include using namespace std; int main() { int n; char cs, ct, cb; while (cin >> n, n) { cin >> cs >> ct >> cb; int s = cs - 'A', t = ct - 'A', b = cb - 'A'; int count[9] = {0}; count[s] = 1; for (int i = 0; i < n; i++) { int plus[9] = {0}; for (int j = 0; j < 9; j++) { // 上に移動 if (j-3 >= 0 && j-3 != b) { plus[j-3] += count[j]; } else { plus[j] += count[j]; } // 下に移動 if (j+3 <= 8 && j+3 != b) { plus[j+3] += count[j]; } else { plus[j] += count[j]; } // 右に移動 if (j+1 <= 8 && j != 2 && j != 5 && j+1 != b) { plus[j+1] += count[j]; } else { plus[j] += count[j]; } // 左に移動 if (j-1 >= 0 && j != 3 && j != 6 && j-1 != b) { plus[j-1] += count[j]; } else { plus[j] += count[j]; } } for (int j = 0; j < 9; j++) { count[j] = plus[j]; } } int sum = 1; for (int i = 0; i < n; i++) sum *= 4; cout << (double)count[t]/sum << endl; } return 0; } #endregion ***Java

Volume10

2013-01-08T16:05:00+09:00

|問題番号|問題名|C|C++|Java| |1000|[[A + B Problem]]|-|○|-| |1001||-|-|-| |1002||-|-|-| |1003||-|-|-| |1004|[[Pair of Primes]]|-|○|-| |1005||-|-|-| |1006||-|-|-| |1007||-|-|-| |1008||-|-|-| |1009||-|-|-| |1010||-|-|-| |1011||-|-|-| |1012||-|-|-| |1013||-|-|-| |1014||-|-|-| |1015||-|-|-| |1016||-|-|-| |1017||-|-|-| |1018||-|-|-| |1019||-|-|-| |1020|[[Cleaning Robot]]|-|○|-| |1021|[[Emacs-like Editor]]|-|-|-| |1022||-|-|-| |1023||-|-|-| |1024||-|-|-| |1025||-|-|-| |1026||-|-|-| |1027|[[A Piece of Cake]]|-|○|-| |1028||-|-|-| |1029||-|-|-| |1030||-|-|-| |1031||-|-|-| |1032||-|-|-| |1033||-|-|-| |1034||-|-|-| |1035||-|-|-| |1036||-|-|-| |1037||-|-|-| |1038||-|-|-| |1039||-|-|-| |1040||-|-|-| |1041||-|-|-| |1042||-|-|-| |1043||-|-|-| |1044||-|-|-| |1045||-|-|-| |1046||-|-|-| |1047||-|-|-| |1048||-|-|-| |1049||-|-|-| |1050||-|-|-| |1051||-|-|-| |1052||-|-|-| |1053||-|-|-| |1054||-|-|-| |1055||-|-|-| |1056||-|-|-| |1057||-|-|-| |1058||-|-|-| |1059||-|-|-| |1060||-|-|-| |1061||-|-|-| |1062||-|-|-| |1063||-|-|-| |1064||-|-|-| |1065||-|-|-| |1066||-|-|-| |1067||-|-|-| |1068||-|-|-| |1069||-|-|-| |1070||-|-|-| |1071||-|-|-| |1072||-|-|-| |1073||-|-|-| |1074||-|-|-| |1075||-|-|-| |1076||-|-|-| |1077||-|-|-| |1078||-|-|-| |1079||-|-|-| |1080||-|-|-| |1081||-|-|-| |1082||-|-|-| |1083||-|-|-| |1084||-|-|-| |1085||-|-|-| |1086||-|-|-| |1087||-|-|-| |1088||-|-|-| |1089||-|-|-| |1090||-|-|-| |1091||-|-|-| |1092||-|-|-| |1093||-|-|-| |1094||-|-|-| |1095||-|-|-| |1096||-|-|-|

Switching Railroad Cars

2013-01-06T17:03:48+09:00

**0013 : Switching Railroad Cars [[問題文>http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0013]] ***解説スタックをプログラムで表現すればいい。 ***プログラム ***C ***C++ #region #include #include using namespace std; int main() { vector cars; int n; while (cin >> n) { if (n != 0) { cars.push_back(n); } else { cout << cars.back() << endl; cars.pop_back(); } } return 0; } #endregion ***Java