题目链接 11 Dimensions Problem Description 11 Dimensions is a cute contestant being talented in math. One day, 11 Dimensions came across a problem but didn't manage to solve it. Today you are taking training here, so 11 Dimensions turns to you for help.…

Fibonacci Again Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 43539 Accepted Submission(s): 20797 Problem Description There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n…

细数一下这两天做过的值得总结的一些题Orz...... HDU 2571 简单dp,但是一开始WA了一发.原因很简单:没有考虑仔细. 如果指向该点的所有点权值都为负数,那就错了(我一开始默认初始值为0) 这是非常基础的典型DAG模型,好久不做,手明显生了…… #include <bits/stdc++.h> using namespace std; #define REP(i,n) for(int i(0); i < (n); ++i) #define rep(i,a,b) for(in…

Design T-Shirt Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6744    Accepted Submission(s): 3167 Problem Description Soon after he decided to design a T-shirt for our Algorithm Board on Free…

Train Problem I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25029    Accepted Submission(s): 9445 Problem Description As the new term comes, the Ignatius Train Station is very busy nowadays…

http://acm.hdu.edu.cn/showproblem.php?pid=1425 题目大意: 给你n个整数,请按从大到小的顺序输出其中前m大的数. 其中n和m都是位于[-500000,500000]. 你说sort?嗯,速度太慢! 是的,水题. sort可以直接过.但是时间不忍直视!500+MS 那么用hash做呗.因为数范围有限且唯一,直接开个bool的数组就好了. 值得一提的是我看别人的代码里面的输入,用自己写的函数,从300+MS到156MS ,当输入大的时候果然输入会成为瓶颈…

Solved Pro.ID Title Ratio(Accepted / Submitted)   1001 Salty Fish 16.28%(7/43)  OK 1002 Nonsense Time 暴力 7.88%(57/723)   1003 Milk Candy 12.90%(4/31)   1004 Speed Dog 26.97%(48/178)   1005 Snowy Smile 8.52%(225/2640)   1006 Faraway 27.92%(98/351)   1…

A.Salty Fish upsolved 题意 偷苹果,每个节点上有\(a[i]\)个苹果,在某些位置有摄像机,看管子树里距离不超过\(k[i]\)的节点,损坏摄像机有\(c[i]\)代价,求最大收益. 做法 听说这题长链剖分,然后我立马去瞅了题解. 妙哉,长链剖分贪心优化最大流 抛开树形结构,这题是个标准的最小割套路题 最小割=最大流,那么考虑在树上贪心的求最大流 自底向上,对于每个摄像机优先流距离更远的节点,这样把更多的机会留给祖先,深度越低,祖先越容易流到 然后长链剖分套个\(map\)…

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链接: http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1005&cid=595 若一个整数的个位数字截去,再从余下的数中,减去个位数的2倍,如果差是7的倍数,则原数能被7整除.如果一次不容易看出,就需要继续上述过程.如6139,过程如下:613-9×2＝595,59-5×2＝49,所以6139是7的倍数. 能被11整除的数的特征把一个数由右边向左边数,将奇位上的数字与偶位上的数字分别加起来,再求它们的差,如果这个差是11的倍…