随笔分类 - 数论,计算几何
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摘要:http://blog.csdn.net/Regina8023/article/details/44949267
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摘要:C - Area Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Description Jerry, a middle school student, addicts himself to m
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摘要:C - Prime number or not Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uSubmit Status Practice FZU 1649DescriptionYour task is sim...
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摘要:poj 1006 题的思路不是很难的,可以转化数学式:现设 num 是下一个相同日子距离开始的天数 p,e,i,d 如题中所设!那么就可以得到三个式子:( num + d ) % 23 == p; ( num + d ) % 28 == e; ( num + d ) % 33 == i;p,e...
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摘要:循环小数 IITime Limit: 1000 MSMemory Limit: 65536 KTotal Submit: 155(55 users)Total Accepted: 92(51 users)Rating: Special Judge: NoDescription对于一个小数,我们记作0...
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摘要:Error CurvesTime Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1198Accepted Submission(s): 460Problem...
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摘要:求第n个数,该数满足至少由3个不同的素数的乘机组成#include#include#include#include#includeusing namespace std;int prim[5000];int ans[5000];int cnt1,cnt2;void getvis(){ cnt...
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摘要:RealPhobiaTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 376Accepted Submission(s): 151Problem De...
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摘要:1、http://acm.hdu.edu.cn/showproblem.php?pid=2440 按照题意知道是一个简单的多边形即凸包,但给出的点并没有按照顺序的,所以需要自己先求出凸包,然后在用随机淬火求费马点。 2、http://acm.hdu.edu.cn/showproblem.php?pi
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摘要:Crossing Rivers Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Problem Descri...
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摘要:相比较ATan,ATan2究竟有什么不同?本篇介绍一下ATan2的用法及使用条件。对于tan(θ) =y/x:θ=ATan(y/x)求出的θ取值范围是[-PI/2, PI/2]。θ=ATan2(y, x)求出的θ取值范围是[-PI, PI]。当 (x,y) 在第一象限, 0 <θ<PI/2.当(x,...
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摘要:#include#include#include#includeconst double pi=3.141592653;int main(){ int t; scanf("%d",&t); while(t--){ int n; scanf("%d",&n); double a...
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摘要:Time limit: 7s Source limit: 50000B Memory limit: 256MBThe first line contains the number of test cases T. T lines follow, one corresponding to each t...
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摘要:RedFieldTime Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 639Accepted Submission(s): 213Problem Desc...
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摘要:欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励)相遇周期Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2465Accepted S...
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摘要:分拆素数和Time Limit: 1000 MSMemory Limit: 32768 KTotal Submit: 176(99 users)Total Accepted: 106(93 users)Rating:Special Judge:NoDescription把一个偶数拆成两个不同素数的和...
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摘要:Power of CryptographyTime Limit:1000MSMemory Limit:30000KTotal Submissions:20371Accepted:10293DescriptionCurrent work in cryptography involves (among ...
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摘要:在三维向量空间中 , 假设a和b是两个向量, 那么它们的叉积c=aXb可如下严格定义。 (1)|c|=|a×b|=|a||b|sin<a,b> (2)c⊥a, 且c⊥b, (3)c的方向要用“右手法则”判断(用右手的四指先表示向量a的方向,然后手指朝着手心的方向摆动到向量b的方向,大拇指所指的方向就
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摘要:ThefigurebelowshowsPascal'sTriangle:BabyHdividesPascal'sTriangleintosomeDiagonals,likethefollowingfigure:BabyHwantstoknowthesumofKnumberinfrontontheMt...
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摘要:对于C(n, m) mod p。这里的n,m,p(p为素数)都很大的情况。就不能再用C(n, m) = C(n - 1,m) + C(n - 1, m - 1)的公式递推了。这里用到Lusac定理For non-negative integersmandnand a primep, the foll...
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