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C - Monkey and Banana **Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u Submit Status Description A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If …
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**Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26534 Accepted Submission(s): 9332 ** Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into …
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I NEED A OFFER! **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18287 Accepted Submission(s): 7320 ** Problem Description Speakless很早就想出国，现在他已经考完了所有需要的考试，准备了所有要准备的材料，于是，便需要去申请学校了。要申请国外的任何大学，你都要交纳一定的申请费用，这可是很惊人的。Speakless没有多少钱，总共只攒了n万美元。他将在m个学校中选择若干的（当然要在他的经济承受范围 …
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饭卡 **Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14225 Accepted Submission(s): 4945 ** Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计，即在购买之前判断余额。如果购买一个商品之前，卡上的剩余金额大于或等于5元，就一定可以购买成功（即使购买后卡上余额为负），否则无法购买（即使金额足够）。所以大家都希望尽量使卡上的余额最少。 某天，食堂中有n种菜出售，每种菜可购买一次。已知每种菜的价格以及卡上的余额，问最 …
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ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 988 Accepted Submission(s): 500 Problem Description Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137". It was on Codefires(CF), an online competitive …
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Wooden Sticks Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 19008 Accepted: 8012 Description There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, …
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简单贪心。 因为填的字母没有次数限制，所以最优策略很容易想到，就是在最后面填最大的。 不用实际去填，算出ans就可以。 1 2 3 4 #include <iostream>5 #include <cmath>6 #include <cstring>7 #include <algorithm>8 9 using namespace std; 10 11 int main() 12 { 13 int k,len; 14 char st[2000]; 15 int a[2000]; 16 memset(a,0,sizeof(a)); 17 int w[50]; 18 …
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简单贪心.... 需要注意的是数据是非负，所以有0的情况要考虑周全，基本都要特殊处理。 多WA了三次，不知道为什么交C++可以过，交G++就不行。 1 2 /* *********************************************** 3Author :111qqz 4Created Time :2016年02月19日 星期五 16时38分28秒 5File Name :code/hdu/1009.cpp 6************************************************ */ 7 8 #include <iostream>9 #include …
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一开始算法想的有点问题。 坑点在于走廊两侧都有房间 也就是说room1和room2对应的位置是一样的 1 to 3 4to6 是没法同时完成的。 做法就是整个扫一遍，看哪个位置的重复次数最大，*10就是答案。 1 2 #include <iostream>3 #include <algorithm>4 #include <cmath>5 #include<cstdio>6 #include <cstring>7 8 using namespace std; 9 10 int main() 11 { 12 int t,n,a[300],b[300]; 13 int …
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题意：求两个相等的圆环的相交的面积.... 简单计算几何+容斥原理？ 扇形面积公式记错调了半天2333333333 这题不难...倒是从学长那里收获了几点关于代码规范的问题... 听说了学长在北京区域赛时把PI定义错了一位结果一直WA的教训.... 以后还是写acos(-1)吧 局部变量和全局变量因为【想怎么其变量名想得整个人都不好了】就起成了一样的...被学长给了差评。 哦，对！还有一个就是发现了cmath库里有一个奇葩的函数名叫y1.。。。。。。。 —————————————————————————————————————————————— 竟然CE了 提示 error:pow(int,int) is ambiguous 看来我 …
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