题目来源:http://poj.org/problem?id=1020 题目大意:有一块边长为s的正方形大蛋糕,有n个客人,每个客人想分一块边长为si的正方形蛋糕.求这块大蛋糕能否恰好满足所有客人的需求而不浪费. 输入:第一行为测试用例数.接下来每行的第一个数位大蛋糕的边长,第二个数位客人的数目n,接下来的n个数为每个客人想要的蛋糕的边长. 输出:若能恰好分完输出“KHOOOOB!”,否则输出“HUTUTU!” Sample Input 2 4 8 1 1 1 1 1 3 1 1 5 6 3 3…

[poj1020]Anniversary Cake Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17203   Accepted: 5619 Description Nahid Khaleh decides to invite the kids of the "Shahr-e Ghashang" to her wedding anniversary. She wants to prepare a square…

Anniversary Cake Time Limit: 1000MSMemory Limit: 10000KB64bit IO Format: %I64d & %I64u Submit Status Description Nahid Khaleh decides to invite the kids of the "Shahr-e Ghashang" to her wedding anniversary. She wants to prepare a square-shap…

                                                                                                              Anniversary Cake            Nahid Khaleh decides to invite the kids of the "Shahr-e Ghashang" to her wedding anniversary. She wants to…

Anniversary Cake Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 15704   Accepted: 5123 Description Nahid Khaleh decides to invite the kids of the "Shahr-e Ghashang" to her wedding anniversary. She wants to prepare a square-shaped c…

http://poj.org/problem?id=1020 (题目链接) 题意 有一个S*S的大蛋糕,还有许多正方形的小蛋糕,问能否将大蛋糕完整的分成所有的小蛋糕,不能有剩余. Solution 像这种看起来很麻烦的基本上都是水题,然而poj上的所谓水题,我也是呵呵了. 在根据题意做完若干判断剪枝后,我们开始搜索.因为蛋糕不能有剩余,所以搜索就很好搜了,刚开始没注意不知道,直接参考了别人的程序→_→.详情请见:题解 代码 // poj1020 #include<algorithm> #inc…

Anniversary Cake Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16579   Accepted: 5403 Description Nahid Khaleh decides to invite the kids of the "Shahr-e Ghashang" to her wedding anniversary. She wants to prepare a square-shaped c…

A. Anniversary Cake 随便挑两个点切掉就好了. #include<bits/stdc++.h> using namespace std; const int Maxn=200020; typedef long long LL; typedef pair<int,int>pi; pi a[2]; int main() { freopen("anniversary.in","r",stdin); freopen("an…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…

BFS广搜题目有时间一个个做下来 2009-12-29 15:09 1574人阅读 评论(1) 收藏 举报 图形graphc优化存储游戏 有时间要去做做这些题目,所以从他人空间copy过来了,谢谢那位大虾啦. pku 1175 Starry Night 题目地址:http://acm.pku.edu.cn/JudgeOnline/problem?id=1175 解法:BFS,要注意的是如何判断图形是一样的,我的做法就是计算每两个点的距离之和. 看:http://hi.baidu.com/doxi…