摘要： Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13897Accepted Submission(s): 4569Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves 阅读全文

摘要： Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13897Accepted Submission(s): 4569Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves 阅读全文

摘要： Just a HookTime Limit: 4000/2000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13564Accepted Submission(s): 6755Problem DescriptionIn the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecu 阅读全文

摘要： Climbing WormTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10288Accepted Submission(s): 6815Problem DescriptionAn inch worm is at the bottom of a well n inches deep. It has enough energy to climb u inches every minute, but then has to rest a minu 阅读全文

摘要： ReversiTime Limit: 5000/2000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1065Accepted Submission(s): 435Problem DescriptionReversi, also called Othello, is a two-sided game.Each of the two sides corresponds to one player; they are referred to here as light and dark 阅读全文

摘要： 树状数组总结问题：已知数组 a[],元素个数为n，现在更改数组中某些元素的值，求更改后a数组中i到j区间内元素的和（1= 1;i -= lowbit(i)) sum += c[i]; return sum;}例如:HDU 1556Color the ballTime Limit: 9000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6412Accepted Submission(s): 3385Problem DescriptionN个气球排成一排，从左... 阅读全文

摘要： Color the ballTime Limit: 9000/3000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6412Accepted Submission(s): 3385Problem DescriptionN个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a #include #include #include #include #include using namespace std;int map[100010];int n;int a 阅读全文

摘要： 如果给定一个数组，要你求里面所有数的和，一般都会想到累加。但是当那个数组很大的时候，累加就显得太耗时了，时间复杂度为O(n)，并且采用累加的方法还有一个局限，那就是，当修改掉数组中的元素后，仍然要你求数组中某段元素的和，就显得麻烦了。所以我们就要用到树状数组，他的时间复杂度为O（lgn），相比之下就快得多。下面就讲一下什么是树状数组： 一般讲到树状数组都会少不了下面这个图： 下面来分析一下上面那个图看能得出什么规律： 据图可知：c1=a1，c2=a1+a2，c3=a3，c4=a1+a2+a3+a4，c5=a5，c6=a5+a6，c7=a7，c8=a1+a2+a3+a4+a5+a6+a7... 阅读全文

摘要： 敌兵布阵Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32178Accepted Submission(s): 13823Problem DescriptionC国的死对头A国这段时间正在进行军事演习，所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务就是要监视这些工兵营地的活动情况。由于采取了某种先进的监测手段，所以每个工兵营地的人数C国都掌握的一清二楚,每个工兵营地的人 阅读全文

摘要： //线段树的节点//节点包括两部分信息，基本域，和信息域//基本域：左右边界leftboundary,rightboundary. 左右孩子：liftchild,rightchild //信息域：value值，如RMQ问题中，信息域中存储的是区间最大值struct Node { int leftboundary,rightboundary; Node *leftchild,*rightchild; typedef value;};//空树的建立，内含value值的初始化；//一般在主函数中首先调用 Node* root= buildtree(1,n);建立一棵新树Node *buildtree 阅读全文