C - Monkey and Banana
**Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u
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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 …
阅读更多**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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**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 …
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**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 …
阅读更多ZCC Loves Codefires
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 988 Accepted Submission(s): 500Problem Description
Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137".
It was on Codefires(CF), an online …
阅读更多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, …
阅读更多简单贪心。 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 …
阅读更多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。
阅读更多题意: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 …
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