35:I[290370,["/_next/static/immutable/chunks/0aneuidq2dd5s.js","/_next/static/immutable/chunks/3asdwddb65r_7.js","/_next/static/immutable/chunks/11qxjuvxxj8-v.js","/_next/static/immutable/chunks/2ol9wyej_s9h-.js","/_next/static/immutable/chunks/3dqr5h54-q26k.js","/_next/static/immutable/chunks/20gnsl3b8ay94.js","/_next/static/immutable/chunks/2v6773b5v9j_y.js","/_next/static/immutable/chunks/437mepgwkzb8b.js","/_next/static/immutable/chunks/03jf6lfc9r-so.js","/_next/static/immutable/chunks/1cty97j_3t4p8.js","/_next/static/immutable/chunks/1atq4abfx_cwi.js","/_next/static/immutable/chunks/29bm85rhaftys.js","/_next/static/immutable/chunks/3vuhqfn8nseb-.js","/_next/static/immutable/chunks/36tubk3midieh.js","/_next/static/immutable/chunks/2ke2tvdiokend.js","/_next/static/immutable/chunks/386n_qt61qywd.js","/_next/static/immutable/chunks/12nj9lbbxts99.js","/_next/static/immutable/chunks/2h62w_fdykycp.js","/_next/static/immutable/chunks/091zjp-oxk1vw.js","/_next/static/immutable/chunks/18bin06et1f2f.js","/_next/static/immutable/chunks/0q669txy1w1q5.js"],"default"] 36:I[218728,["/_next/static/immutable/chunks/0aneuidq2dd5s.js","/_next/static/immutable/chunks/3asdwddb65r_7.js","/_next/static/immutable/chunks/11qxjuvxxj8-v.js","/_next/static/immutable/chunks/2ol9wyej_s9h-.js","/_next/static/immutable/chunks/3dqr5h54-q26k.js","/_next/static/immutable/chunks/20gnsl3b8ay94.js","/_next/static/immutable/chunks/2v6773b5v9j_y.js","/_next/static/immutable/chunks/437mepgwkzb8b.js","/_next/static/immutable/chunks/03jf6lfc9r-so.js","/_next/static/immutable/chunks/1cty97j_3t4p8.js","/_next/static/immutable/chunks/1atq4abfx_cwi.js","/_next/static/immutable/chunks/29bm85rhaftys.js","/_next/static/immutable/chunks/3vuhqfn8nseb-.js","/_next/static/immutable/chunks/36tubk3midieh.js","/_next/static/immutable/chunks/2ke2tvdiokend.js","/_next/static/immutable/chunks/386n_qt61qywd.js","/_next/static/immutable/chunks/12nj9lbbxts99.js","/_next/static/immutable/chunks/2h62w_fdykycp.js","/_next/static/immutable/chunks/091zjp-oxk1vw.js","/_next/static/immutable/chunks/18bin06et1f2f.js","/_next/static/immutable/chunks/0q669txy1w1q5.js"],"default"] 37:T4cc,Suppose that $1$ is at position $k$ in a good permutation $p$. The maximum of $p_{k-1}$ and $1$ is $p_{k-1}$, so $p_{k-1}=b_{k-1}$. Moving to the left, the values of $b$ are distinct, so the maximum on each next edge must be its left endpoint. Therefore $p_i=b_i$ for every $ik$. Thus, every possible $p$ is obtained by inserting $1$ somewhere into $b$. ## Where Can We Insert $1$? To the left of $1$, every adjacent maximum must be the left element, so that part of $b$ must be strictly decreasing. To the right of $1$, every adjacent maximum must be the right element, so that part must be strictly increasing. The minimum element of $b$ is $2$. Consequently, the boundary between the decreasing and increasing parts must be next to $2$. There are only two candidates: inserting $1$ immediately before $2$, or immediately after $2$. Let $j$ be the position of $2$ in $b$. A solution exists exactly when $$ b_1>b_2>\cdots>b_j=2\n#include \"testlib.h\"\n\nusing namespace std;\n\nint main(int argc, char* argv[]) {\n registerTestlibCmd(argc, argv);\n\n int n = inf.readInt();\n int k = inf.readInt();\n\n string answer = ouf.readToken();\n for (char& c : answer) {\n c = static_cast(tolower(static_cast(c)));\n }\n\n if (answer != \"yes\" && answer != \"no\") {\n quitf(_wa, \"expected Yes or No\");\n }\n\n bool possible = (n - k != 1);\n if ((answer == \"yes\") != possible) {\n quitf(_wa, \"wrong answer\");\n }\n\n if (!ouf.seekEof()) {\n quitf(_wa, \"extra tokens in the output\");\n }\n\n quitf(_ok, \"correct\");\n}\n"},"solvedCount":100,"attemptedCount":103,"tags":["math"],"learnPages":[{"title":"Who Must Move?","content":"Suppose exactly $k$ people receive their own gifts. Choose these people and give each of them their own gift.\n\nThere are now $m=n-k$ people and their $m$ gifts left. None of these people may receive their own gift, so we need to arrange the remaining gifts without any fixed point."},{"title":"The Only Impossible Size","content":"If $m=0$, all gifts have already been distributed, so the answer is possible.\n\nIf $m=1$, the only remaining person must receive the only remaining gift, which is their own. This would create one more correct gift, so the answer is impossible.\n\nFor every $m \\ge 2$, place the remaining people in any order and give each person the gift belonging to the next person, with the last person receiving the first person's gift. Everyone receives a different person's gift, so the answer is possible."},{"title":"Algorithm","content":"The answer is `No` exactly when $n-k=1$. In every other case, the answer is `Yes`.\n\nThe time complexity is $O(1)$ and the memory usage is $O(1)$."}],"hasLearnContent":true,"likeCount":39,"neutralCount":0,"dislikeCount":9,"isInteractive":false,"interactor":"$undefined","hidden":false,"sampleTestCases":"$undefined","samplesDiverged":"$undefined","sourceAttribution":"$undefined","source":"$undefined","validatorSource":"$undefined","correctSolution":"$undefined","generators":"$undefined","genDataContent":"$undefined","readiness":"$undefined","linkedBuildSessionId":"$undefined","isLinked":"$undefined","authors":[{"_id":"692eca62f11e323ac6f94d25","name":"AmirMohammad Shahrezaei (AmShZ)"}],"statementOverrides":"$undefined","createdAt":"2026-08-29T18:27:15.767Z","updatedAt":"2026-09-04T06:37:54.806Z"},{"_id":"6a9326b8ff44cebdb9de97e7","contestId":"6a72c8e20f72a03f001a6408","slug":"adjacent-maximums","title":"Adjacent Maximums","description":"You are given a permutation $b$ of the integers from $2$ to $n$.\n\nA permutation $p$ of the integers from $1$ to $n$ is called **good** if, for every $1 \\le i < n$,\n\n$$\nb_i = \\max(p_i,p_{i+1}).\n$$\n\nFind the number of good permutations.","constraints":"- $2 \\le n \\le 1000$\n- $b$ is a permutation of the integers from $2$ to $n$","difficulty":"easy","order":5,"label":"B","releaseDay":"$undefined","points":1000,"vivePrize":"$undefined","submissionKind":"code","proofJudgeTier":"$undefined","proofJudgeCount":"$undefined","proofSubmissionMode":"$undefined","proofLeanStatement":"$undefined","answerFormat":"$undefined","testCases":[{"input":"6\n6 4 2 3 5\n","expectedOutput":"2\n","isSample":true,"hidden":false,"explanation":"The two good permutations are $[6,4,1,2,3,5]$ and $[6,4,2,1,3,5]$. For each of them, the maximums of adjacent pairs are $[6,4,2,3,5]$.\n"},{"input":"5\n2 5 3 4\n","expectedOutput":"0\n","isSample":true,"hidden":false,"explanation":"The first required maximum forces the first two elements of $p$ to be $1$ and $2$ in some order. The second required maximum then forces the third element to be $5$, making the next adjacent maximum at least $5$ instead of $3$. Therefore, no good permutation exists.\n"}],"sampleCount":2,"inputFormat":"The input consists of two lines:\n\n$$\nn\n$$\n\n$$\nb_1 \\quad b_2 \\quad \\cdots \\quad b_{n-1}\n$$","outputFormat":"Output one integer: the number of good permutations.","interactionFormat":"$undefined","timeLimit":1,"memoryLimit":256,"checker":{"type":"wcmp"},"solvedCount":91,"attemptedCount":94,"tags":["math"],"learnPages":[{"title":"Recovering the Permutation","content":"$37"}],"hasLearnContent":true,"likeCount":36,"neutralCount":0,"dislikeCount":5,"isInteractive":false,"interactor":"$undefined","hidden":false,"sampleTestCases":"$undefined","samplesDiverged":"$undefined","sourceAttribution":"$undefined","source":"$undefined","validatorSource":"$undefined","correctSolution":"$undefined","generators":"$undefined","genDataContent":"$undefined","readiness":"$undefined","linkedBuildSessionId":"$undefined","isLinked":"$undefined","authors":[{"_id":"692eca62f11e323ac6f94d25","name":"AmirMohammad Shahrezaei (AmShZ)"}],"statementOverrides":"$undefined","createdAt":"2026-08-29T18:36:40.201Z","updatedAt":"2026-09-04T06:37:54.867Z"},{"_id":"6a9330c3ff44cebdb9de97ed","contestId":"6a72c8e20f72a03f001a6408","slug":"one-way-parking","title":"One-Way Parking","description":"There are $n$ parking spaces arranged in a circle and numbered from $1$ to $n$ in clockwise order. There are also $n$ cars, numbered from $1$ to $n$.\n\nCar $i$ enters the parking area at space $a_i$. It may park there or continue moving clockwise until it reaches another space. Cars cannot move counterclockwise.\n\nBefore any car enters, you must assign a distinct parking space to every car. If a car parks at its entry space, its cost is $0$. Each move from one space to the next space clockwise increases its cost by $1$.\n\nFind the minimum possible value of the maximum cost among all cars.","constraints":"- $1 \\le n \\le 2 \\times 10^5$\n- $1 \\le a_i \\le n$","difficulty":"medium","order":6,"label":"D","releaseDay":"$undefined","points":2000,"vivePrize":"$undefined","submissionKind":"code","proofJudgeTier":"$undefined","proofJudgeCount":"$undefined","proofSubmissionMode":"$undefined","proofLeanStatement":"$undefined","answerFormat":"$undefined","testCases":[{"input":"4\n1 2 3 4\n","expectedOutput":"0\n","isSample":true,"hidden":false,"explanation":"Assign every car to its entry space. Every car then has cost $0$.\n"},{"input":"5\n1 1 1 1 1\n","expectedOutput":"4\n","isSample":true,"hidden":false,"explanation":"The five cars can be assigned to spaces $1,2,3,4,5$, giving costs $0,1,2,3,4$. Since all cars enter at the same space and must receive distinct spaces, no smaller maximum cost is possible.\n"},{"input":"7\n2 2 3 3 5 5 6\n","expectedOutput":"2\n","isSample":true,"hidden":false,"explanation":"Assign the cars, in input order, to spaces $2,3,4,5,6,7,1$. Their costs are $0,1,1,2,1,2,2$.\n\nA maximum cost of $1$ is impossible: the four cars entering at spaces $2$ and $3$ could then only use spaces $2,3,4$.\n"}],"sampleCount":3,"inputFormat":"The input consists of:\n\n$$\n\\begin{array}{l}\nn \\\\\na_1 \\quad a_2 \\quad \\cdots \\quad a_n\n\\end{array}\n$$","outputFormat":"The output consists of:\n\n$$\nD\n$$\n\nwhere $D$ is the minimum possible value of the maximum cost among all cars.","interactionFormat":"$undefined","timeLimit":2,"memoryLimit":256,"checker":{"type":"wcmp"},"solvedCount":38,"attemptedCount":65,"tags":["math","greedy"],"learnPages":[{"title":"Counting Cars at Each Entrance","content":"$38"}],"hasLearnContent":true,"likeCount":21,"neutralCount":0,"dislikeCount":0,"isInteractive":false,"interactor":"$undefined","hidden":false,"sampleTestCases":"$undefined","samplesDiverged":"$undefined","sourceAttribution":"$undefined","source":"$undefined","validatorSource":"$undefined","correctSolution":"$undefined","generators":"$undefined","genDataContent":"$undefined","readiness":"$undefined","linkedBuildSessionId":"$undefined","isLinked":"$undefined","authors":[{"_id":"692eca62f11e323ac6f94d25","name":"AmirMohammad Shahrezaei (AmShZ)"}],"statementOverrides":"$undefined","createdAt":"2026-08-29T19:19:31.954Z","updatedAt":"2026-09-04T06:37:55.053Z"},{"_id":"6a933147ff44cebdb9de97f2","contestId":"6a72c8e20f72a03f001a6408","slug":"final-cut","title":"Final Cut","description":"A movie consists of $n$ scenes shown in order. Scene $i$ features actor $a_i$.\n\nThe final cut is called **smooth** if no two consecutive scenes feature the same actor.\n\nThe editor may delete at most one non-empty contiguous segment of scenes. The remaining scenes keep their relative order. The editor may also choose not to delete anything.\n\nFind the minimum number of scenes that must be deleted to make the final cut smooth.","constraints":"- $1 \\le n \\le 2 \\times 10^5$\n- $1 \\le a_i \\le 10^9$","difficulty":"easy","order":7,"label":"C","releaseDay":"$undefined","points":1500,"vivePrize":"$undefined","submissionKind":"code","proofJudgeTier":"$undefined","proofJudgeCount":"$undefined","proofSubmissionMode":"$undefined","proofLeanStatement":"$undefined","answerFormat":"$undefined","testCases":[{"input":"7\n1 2 1 3 2 3 1\n","expectedOutput":"0\n","isSample":true,"hidden":false,"explanation":"No two consecutive scenes feature the same actor, so no scene needs to be deleted.\n"},{"input":"7\n1 2 2 3 4 4 2\n","expectedOutput":"3\n","isSample":true,"hidden":false,"explanation":"Delete scenes $3$ through $5$. The remaining actor sequence is $1,2,4,2$, which is smooth. Deleting fewer than three scenes cannot eliminate both repeated adjacent pairs with one contiguous deletion.\n"},{"input":"7\n1 3 3 2 3 3 1\n","expectedOutput":"4\n","isSample":true,"hidden":false,"explanation":"Delete scenes $2$ through $5$. The remaining actor sequence is $1,3,1$, which is smooth. Any contiguous deletion of at most three scenes leaves two equal actors next to each other.\n"}],"sampleCount":3,"inputFormat":"The input consists of two lines:\n\n$$\nn\n$$\n\n$$\na_1 \\quad a_2 \\quad \\cdots \\quad a_n\n$$","outputFormat":"Output one integer: the minimum number of scenes that must be deleted.","interactionFormat":"$undefined","timeLimit":1,"memoryLimit":256,"checker":{"type":"wcmp"},"solvedCount":83,"attemptedCount":83,"tags":["math"],"learnPages":[{"title":"Conflicting Neighbors","content":"$39"}],"hasLearnContent":true,"likeCount":37,"neutralCount":0,"dislikeCount":1,"isInteractive":false,"interactor":"$undefined","hidden":false,"sampleTestCases":"$undefined","samplesDiverged":"$undefined","sourceAttribution":"$undefined","source":"$undefined","validatorSource":"$undefined","correctSolution":"$undefined","generators":"$undefined","genDataContent":"$undefined","readiness":"$undefined","linkedBuildSessionId":"$undefined","isLinked":"$undefined","authors":[{"_id":"692eca62f11e323ac6f94d25","name":"AmirMohammad Shahrezaei (AmShZ)"}],"statementOverrides":"$undefined","createdAt":"2026-08-29T19:21:43.411Z","updatedAt":"2026-09-04T06:37:54.930Z"}],"contestId":"6a72c8e20f72a03f001a6408","contestUrlId":"21","isAdmin":false,"isTester":false,"isStarted":true,"isRegistered":false,"canRegister":false,"problemStatusMap":{},"isPremiumBlocked":false,"sponsorshipSnapshot":"$1a:props:children:1:props:children:0:props:sponsorshipSnapshot","recentRoundParticipantCounts":[92,108,96],"participationMode":null,"virtualStartTime":null,"contestDurationMs":7200000}]]}]

ARight Gifts	500	100/103
BAdjacent Maximums	1000	91/94
CFinal Cut	1500	83/83
DOne-Way Parking	2000	38/65
EBig Pick	2500	37/42
FSorting the Pieces	3000	23/25
GTree Broadcast	3500	9/11