随笔分类 - 算法竞赛
摘要:第一道是poj3020.题意: 另一篇博客的解释: Antenna Placement Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu Submit Status Practice POJ 3020 Descrip
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摘要:Snacks Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1295 Accepted Submission(s): 302 Problem
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摘要:void print() { int j=1,t=1; for(int i=1; ipow(2,(t-1))) { printf("\n"); j=1; t++; } } printf("\n\n"); }
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摘要:Teacher Bo Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1014 Accepted Submission(s): 561 Pro
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摘要:A Simple Problem with Integers Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%lld & %llu Submit Status Practice POJ 3468 A Simple Problem wi
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摘要:I Hate It Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1754 I Hate It Submit Status Practice HDU 17
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摘要:#include #include #include using namespace std; const int maxx = 200010; int tree[maxx>1; build(root>1; if(pos>1; if(qrmid) { return query((rootpow(2,(t-1))) { ...
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摘要:敌兵布阵 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1166 敌兵布阵 Submit Status Practice HDU 1166 Descrip
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摘要:1010 Rower Bo 首先这个题微分方程强解显然是可以的,但是可以发现如果设参比较巧妙就能得到很方便的做法。 先分解v_1v1, 设船到原点的距离是rr,容易列出方程 \frac{ dr}{ dt}=v_2\cos \theta-v_1dtdr=v2cosθ−v1
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摘要:这里是一个比较简单的问题:考虑每个数对和的贡献。先考虑数列两端的值,两端的摆放的值总计有2种,比如左端:0,大,小;0,小,大;有1/2的贡献度。右端同理。 中间的书总计有6种可能。小,中,大。其中有两种对答案有贡献,即1/3的贡献度。加和计算可得到答案。 Permutation Bo Time L
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摘要:Sqrt Bo Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 980 Accepted Submission(s): 452 Problem
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摘要:迷宫城堡 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12778 Accepted Submission(s): 5698 Problem D
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摘要:Sticks Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1455 Sticks Submit Status Practice HDU 1455 Des
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摘要:变形课 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 21512 Accepted Submission(s): 7761 Problem D
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摘要:D - Oil Deposits Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1241 D - Oil Deposits Submit Status P
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摘要:如果所给的时间(步数) t 小于最短步数path,那么一定走不到。 若满足t>path。但是如果能在恰好 t 步的时候,走到出口处。那么(t-path)必须是二的倍数。 关于第二种方案的解释: 这种方案学名为“奇偶剪枝”。我们已知了最短的步数就是直角三角形的两条直角边,实际上的路径却不一定非要沿着这
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摘要:B - Prime Ring Problem Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1016 B - Prime Ring Problem Sub
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摘要:Square Coins Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1398 Square Coins Submit Status Practice
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