09 2016 档案
摘要:树状数组 数据结构知识点1-树状数组 树状数组的用途就是维护一个数组,重点不是这个数组,而是要维护的东西,最常用的求区间和问题,单点更新。但是某些大牛YY出很多神奇的东西,完成部分线段树能完成的功能,比如区间更新,区间求最值问题。 树状数组当然是跟树有关了,但是这个树是怎么构建的呐?这里就不得不感叹
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摘要:Description N(3N20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To
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摘要:Turing Tree Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4671 Accepted Submission(s): 1649 Pro
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摘要:Cube Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 1949 Accepted Submission(s): 1013 Problem D
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摘要:See you~ Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 4753 Accepted Submission(s): 1518 Proble
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摘要:A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5402 Accepted Subm
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摘要:这个题刚开始的时候是套模板的,并没有真的理解什么树状数组的区间更新,这几天想了一下,下面是总结: 区间更新这里引进了一个数组delta数组,delta[i]表示区间 [i, n] 的共同增量,每次你需要更新的时候只需要更新delta数组就行了,因为每段区间更新的数都记录在这个数组中,那怎么查询前缀和
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摘要:这个求一千亿以内的素数大约用6780ms 这个完全就是爆炸了 #include <bits/stdtr1c++.h> #define MAXN 100 // pre-calc max n for phi(m, n) #define MAXM 100010 // pre-calc max m for
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摘要:A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 97217 Accepted: 30358 Case Time Limit: 2000MS Description Y
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摘要:Flowers Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3150 Accepted Submission(s): 1549 Problem
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摘要:敌兵布阵 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 78528 Accepted Submission(s): 33171 Problem
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摘要:/* 因为归并排序至于O(n)的复杂度所以省时 */ #include #include #include #include #include #include #include #define N 100010 #define MAXN 100010 using namespace std; //将有二个有序数列a[first...mid]和a[mid...last]合并。 void m...
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摘要:Color the ball Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17276 Accepted Submission(s): 8640
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摘要:Football Games Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 5055 Accepted Submission(s): 904 P
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摘要:Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 42707 Accepted: 18587 Description Astronomers often examine star maps where stars are
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摘要:Ultra-QuickSort Time Limit: 7000MS Memory Limit: 65536K Total Submissions: 56054 Accepted: 20706 Description In this problem, you have to analyze a pa
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摘要:首先人脑能识别的,也就是我们常写的就是原码,因为数字都有正负之分所以,二进制的时候用最高位表示正负,0为正,1为负,例如3二进制位00000011,-3二进制位10000011. 反码:正数的反码等于其本身,负数的反码:符号位不变其他位取反。(00000011)原=(00000011)反,(1000
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摘要:Problem A: 2016 Description 给出正整数 n 和 m,统计满足以下条件的正整数对 (a,b) 的数量: 1. 1≤a≤n,1≤b≤m; 2. a×b 是 2016 的倍数。 给出正整数 n 和 m,统计满足以下条件的正整数对 (a,b) 的数量: 1. 1≤a≤n,1≤b≤
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摘要:Sliding Window Time Limit: 12000MS Memory Limit: 65536K Total Submissions: 54824 Accepted: 15777 Case Time Limit: 5000MS Description An array of size
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