随笔分类 - 概率DP
ZOJ 3822(求期望)
摘要:DominationTime Limit:8 Seconds Memory Limit:131072 KB Special JudgeEdward is the headmaster of Marjar University. He is enthusiastic about chess and o...
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LOOPS(HDU 3853)
摘要:LOOPSTime Limit: 15000/5000 MS (Java/Others)Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 3163Accepted Submission(s): 1279Problem Des...
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Card Collector(HDU 4336)
摘要:Card CollectorTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3001Accepted Submission(s): 1435Spec...
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Help Me Escape (ZOJ 3640)
摘要:J - Help Me EscapeCrawling in process... Crawling failed Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %lluSubmit Status Practice ZOJ ...
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Collecting Bugs(POJ 2096)
摘要:Collecting BugsTime Limit:10000MSMemory Limit:64000KTotal Submissions:3064Accepted:1505Case Time Limit:2000MSSpecial JudgeDescriptionIvan is fond of c...
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Kids and Prizes(SGU 495)
摘要:495. Kids and PrizesTime limit per test: 0.25 second(s)Memory limit: 262144 kilobytesinput: standardoutput: standardICPC (International Cardboard Prod...
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Check the difficulty of problems(POJ 2151)
摘要:Check the difficulty of problemsTime Limit:2000MSMemory Limit:65536KTotal Submissions:5457Accepted:2400DescriptionOrganizing a programming contest is ...
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Bag of mice(CodeForces 148D )
摘要:D. Bag of micetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe dragon and the princess are ar...
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Aeroplane chess(HDU 4405)
摘要:Aeroplane chessTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2060Accepted Submission(s): 1346Pro...
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Football(POJ3071)
摘要:FootballTime Limit:1000MSMemory Limit:65536KTotal Submissions:3469Accepted:1782DescriptionConsider a single-elimination football tournament involving ...
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Scout YYF I(POJ 3744)
摘要:Scout YYF ITime Limit:1000MSMemory Limit:65536KTotal Submissions:5565Accepted:1553DescriptionYYF is a couragous scout. Now he is on a dangerous missio...
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快速幂(整数 + 矩阵)
摘要:一、整数m ^ n % k 的快速幂:ll quickpow(ll m, ll n , ll k){ ll ans = 1; while(n){ if(n & 1)//如果n是奇数 ans = (ans * m) % k; n = n...
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递推式转化为矩阵形式
摘要:EXAMPLE: 递推式: d(n + 2) = p * d(n + 1) + (1 - p) * d(n); 令G(n) = (d(n + 2), d(n + 1))^T; 则 G(n + 1) = M * G(n); 解得 M = p 1 - p 1 0 G(n) = (M ^ ...
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