龙崎的专栏

Language:DefaultBlocksTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3714 Accepted: 1643DescriptionPanda has received an assignment of painti

E - Stocks PredictionTime Limit:8000MS     Memory Limit:0KB     64bit IO Format:%lld & %lluSubmit Status Practice SPOJ AMR10EDescriptionThe department store where my family goes

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

扩展KMP：给出模板串A和子串B，长度分别为lenA和lenB，要求在线性时间内，对于每个A[i]（0

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

GTY's birthday giftTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 307    Accepted Submission(s): 111Problem DescriptionFFZ's birthd

B. ZgukistringZProfessor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.GukiZ has strings a, b, and c. He wants to obtain stringk by

MagicianProblem DescriptionFantasy magicians usually gain their ability through one of three usual methods: possessing it as an innate talent, gaining it through study and practice, or receiving it fro

思路：a[1] + a[2] + a[3] + ... a[n]= (1+a[n/2])(a[1]+...+a[n/2])

p=a+b;     q=a*b;          f[0]=2;     f[1]=a+b;     f[2]=a^2+b^2=(a+b)*(a+b)-2*a*b=(a+b)*f[1]-a*b*f[0];     ...     f[n]=a^n+b^n=(a^(n-1)+b^(n-1))*(a+b)-a*b*(a^(n-2)+b^(n-2))=p*f[n-1]-q*f[n-2]

这个不是直接求矩阵快速幂，是求的和，但可以转化成快速幂。参见《挑战程序设计竞赛》。下面是代码：#include#include#include#include#include#include#include#include#include#includetypedef long long LL;using namespace std;int n,m,k;t

Language:DefaultPseudoprime numbersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5886 Accepted: 2345DescriptionFermat's theorem states that fo

Language:DefaultRaising Modulo NumbersTime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 4103 Accepted: 2335DescriptionPeople are different. Some s

A - Puzzles#include#includeusing namespace std;const int maxn=60;const int INF=1000000000;int n,m,a[maxn],sum[maxn];int main(){ cin>>n>>m; for(int i=1;i<=m;i++) cin>>a[i];

http://acm.whu.edu.cn/land/problem/detail?problem_id=1540&contest_id=12斐波纳契数立方和公式但是除取模再除2不对，然后yy一下，对MOD的两倍取模再除2就对了#include#include#include#include#includeusing namespace std;typedef long

构造出的矩阵：

详见训练指南

根据题意可以构造出矩阵：A

A - 1001DescriptionAn Arc of Dream is a curve defined by following function:AoD(n)=∑n−1i=0ai∗biAoD(n)=\sum_{i=0}^{n-1}a_i*b_iwhere a0=A0a_0 = A_0 ai=ai−1∗AX+AYa_i = a_{i-1}*AX+AY b0=B0b_0 = B_0 bi=