{"id":1310,"date":"2014-10-02T12:50:49","date_gmt":"2014-10-02T04:50:49","guid":{"rendered":"http:\/\/blog.dayandcarrot.net\/?p=1310"},"modified":"2014-10-02T12:50:49","modified_gmt":"2014-10-02T04:50:49","slug":"pat-1012-the-best-rank-25","status":"publish","type":"post","link":"https:\/\/dayandcarrot.space\/?p=1310","title":{"rendered":"PAT 1012. The Best Rank (25)"},"content":{"rendered":"

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C – C Programming Language, M – Mathematics (Calculus or Linear Algebra), and E – English. At the mean time, we encourage students by emphasizing on their best ranks — that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.
\nFor example, The grades of C, M, E and A – Average of 4 students are given as the following:<\/p>\n

StudentID  C  M  E  A\n310101     98 85 88 90\n310102     70 95 88 84\n310103     82 87 94 88\n310104     91 91 91 91\n<\/pre>\n
Then the best ranks for all the students are No.1<\/i> since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.
\nInput<\/b>
\nEach input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.
\nOutput<\/b>
\nFor each of the M students, print in one line the best rank for him\/her, and the symbol of the corresponding rank, separated by a space.
\nThe priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.
\nIf a student is not on the grading list, simply output “N\/A”.
\nSample Input<\/b><\/p>\n
5 6\n310101 98 85 88\n310102 70 95 88\n310103 82 87 94\n310104 91 91 91\n310105 85 90 90\n310101\n310102\n310103\n310104\n310105\n999999\n<\/pre>\n
Sample Output<\/b><\/p>\n
1 C\n1 M\n1 E\n1 A\n3 A\nN\/A<\/pre>\n
===============================================<\/p><\/blockquote>\n
\n
\u53c8\u56de\u6765\u505aPAT\u9898\u76ee\u4e86\uff0cC++\u90fd\u5fd8\u5f97\u5dee\u4e0d\u591a\u5e72\u51c0\u4e86\u6709\u6ca1\u6709\uff01\u4e0b\u6b21\u5c1d\u8bd5\u4e0b\u63d0\u4ea4Python\u7684\u7b54\u6848\u5427~\u8bf6\u563f\u563f
\n\u8fd9\u9053\u9898\u6ca1\u4ec0\u4e48\u96be\u7684\uff0c\u56e0\u4e3a\u5b83\u65f6\u95f4\u8981\u6c42\u4e0d\u9ad8\u3002
\n\u6211\u505a\u4e86\u4e2a\u94fe\u8868\u628a\u5b66\u751f\u6210\u7ee9\u7684\u8282\u70b9\u8fde\u8d77\u6765\uff0c\u8fd9\u6837\u662f\u56e0\u4e3a\u5b66\u53f7\u67096\u4f4d\u6570\uff08\u5176\u5b9e\u8981\u662f\u5f04\u4e2a1000000\u884c\u7684\u77e9\u9635\u4e5f\u6ca1\u4ec0\u4e48\u4e86\u4e0d\u8d77\u7684\u561b\u54fc\uff01\uff09\uff0c\u53cd\u6b63\u4e4b\u540e\u5c31\u662f\u94fe\u8868\u67e5\u627e\u7684\u95ee\u9898\u4e86\uff0c\u6ca1\u5565\u53ef\u8bf4\u3002<\/p>\n
#include <iostream>\nusing namespace std;\nstruct RecordNode\n{\n\t\/\/\u5b58\u653e\u5355\u4e2a\u5b66\u751f\u6570\u636e\u7684\u94fe\u8868\u8282\u70b9\n\tint studentID;\n\tint results[4];\t\/\/0-A, 1-C, 2-M, 3-E\n\tRecordNode* nextNode;\n\tRecordNode(int sid, int c, int m, int e, int a)\n\t{\n\t\tstudentID = sid;\n\t\tresults[0] = a;\n\t\tresults[1] = c;\n\t\tresults[2] = m;\n\t\tresults[3] = e;\n\t\tnextNode = NULL;\n\t}\n};\nRecordNode* rootNode = NULL;\nRecordNode* getNodePointer(int studentID)\n{\n\t\/\/\u6ca1\u6709\u5219\u8fd4\u56deNULL\n\tRecordNode* currNode = rootNode;\n\twhile (currNode != NULL)\n\t{\n\t\tif (currNode->studentID == studentID)\n\t\t\treturn currNode;\n\t\tcurrNode = currNode->nextNode;\n\t}\n\treturn currNode;\n}\nvoid getRankInMatrix(int studentID, int* ranks)\n\/\/\u4ece\u77e9\u9635\u4e2d\u5f97\u5230\u67d0\u4eba\u7684all\u6392\u4f4d\n\/\/(subj:  0-C, 1-M, 2-E, 3-A)\n{\n\t\/\/int ranks[4] = { 1, 1, 1, 1 };\n\tRecordNode* node = getNodePointer(studentID);\n\tif (node == NULL)\n\t{\n\t\t\/\/ranks = NULL;\n\t\t\/\/delete ranks;\n\t\tranks[0] = -1;\n\t\treturn;\n\t}\n\tint* results = node->results;\t\/\/\u6240\u6709\u8bfe\u7684\u6210\u7ee9\n\tRecordNode* opNode = rootNode;\n\twhile (opNode != NULL)\n\t{\n\t\t\/\/\u904d\u5386\n\t\tfor (int subj = 0; subj < 4; subj++)\n\t\t{\n\t\t\t\/\/subjects\n\t\t\tif (opNode->results[subj] > results[subj])\n\t\t\t\tranks[subj] ++;\n\t\t}\n\t\topNode = opNode->nextNode;\n\t}\n\t\/\/return ranks;\n}\nchar ACME[] = \"ACME\";\nint main()\n{\n\tint M, N;\n\tcin >> N >> M;\n\tRecordNode* opNode = rootNode;\n\tfor (int i = 0; i < N; i++)\n\t{\n\t\tint sid, c, m, e, a;\n\t\tcin >> sid >> c >> m >> e;\n\t\ta = (int)(((c + m + e) \/ 3.0) + 0.5); \/\/\u56db\u820d\u4e94\u5165\n\t\tif (opNode == NULL)\n\t\t{\n\t\t\t\/\/\u65b0\u5efa\u6839\u8282\u70b9\n\t\t\topNode = new RecordNode(sid, c, m, e, a);\n\t\t\trootNode = opNode;\n\t\t\tcontinue;\n\t\t}\n\t\tRecordNode *currNode = new RecordNode(sid, c, m, e, a);\n\t\topNode->nextNode = currNode;\n\t\topNode = opNode->nextNode;  \/\/move forward\n\t}\n\tfor (int i = 0; i < M; i++)\n\t{\n\t\tint sid;\n\t\tcin >> sid;\n\t\tint ranks[] = { 1, 1, 1, 1 };\n\t\tgetRankInMatrix(sid, ranks);\n\t\tif (ranks[0] == -1)\n\t\t{\n\t\t\tcout << \"N\/A\";\n\t\t\tif (i < M - 1)\n\t\t\t\tcout << endl;\n\t\t\tcontinue;\n\t\t}\n\t\tint bestRank = ranks[0];\n\t\tint bestRankSubj = 0;\n\t\tfor (int x = 1; x < 4; x++)\n\t\t{\n\t\t\tif (ranks[x] < bestRank)\n\t\t\t{\n\t\t\t\tbestRank = ranks[x];\n\t\t\t\tbestRankSubj = x;\n\t\t\t}\n\t\t}\n\t\tcout << bestRank << \" \" << ACME[bestRankSubj];\n\t\tif (i < M - 1)\n\t\t\tcout << endl;\n\t}\n\treturn 0;\n}<\/pre>\n
 
\n <\/p>\n","protected":false},"excerpt":{"rendered":"
To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C – C Programming Language, M – Mathematics (Calculus or Linear Algebra), and E – English. At the mean time, we encourage students by emphasizing on their best ranks — that is, among the four ranks […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[84],"class_list":["post-1310","post","type-post","status-publish","format-standard","hentry","category-study","tag-pat"],"_links":{"self":[{"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=\/wp\/v2\/posts\/1310","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1310"}],"version-history":[{"count":0,"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=\/wp\/v2\/posts\/1310\/revisions"}],"wp:attachment":[{"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1310"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1310"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/dayandcarrot.space\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1310"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}