摘要： Magic BraceletTime Limit: 2000MSMemory Limit: 131072KTotal Submissions: 2537Accepted: 844DescriptionGinny’s birthday is coming soon. Harry Potter is preparing a birthday present for his new girlfriend. The present is a magic bracelet which consists of n magic beads. The are m kinds of different magi 阅读全文

摘要： http://acmicpc.info/archives/505 阅读全文

摘要： ColorTime Limit: 2000MSMemory Limit: 65536KTotal Submissions: 4295Accepted: 1437DescriptionBeads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the neck 阅读全文

摘要： Catalan数 中文:卡特兰数 原理： 令h(1)＝1,h(0)=1，catalan数满足递归式： h(n)= h(1)*h(n-1) + h(2)*h(n-2) + ... + h(n-1)h(1) (其中n>=2) 另类递归式： h(n)=((4*n-2)/(n+1))*h(n-1); 该递推关系的解为： h(n)=C(2n,n)/(n+1) (n=1,2,3,...) 我并不关心其解是怎么求出来的，我只想知道怎么用catalan数分析问题。 我总结了一下，最典型的四类应用：（实质上却都一样，无非是递归等式的应用，就看你能不能分解问题写出递归式了） 1.括号化问题。... 阅读全文

摘要： Epai={pai[v],v:v-V&&pai[v]!=NIL} Epai为树边；深搜对每个顶点都加了一个时间戳，每个顶点有两个时间戳，当顶点v第一次被发现（被标记成灰色）时，记下第一个时间戳的d[v],当结束检查v的邻接表（黑色），记下第二个时戳f[v],d[v]<f[v],顶点v在d[v]之前是白色，在d[v]和f[v]之间为灰色，以后就变成黑色了；DFS(G)for each vertex u-v[G] do color-white pai[v]=NIL time=0 for each u-v[G] do if color[u]=white dfs(u)dfs_vi 阅读全文

摘要： Anti-prime SequencesTime Limit: 3000MSMemory Limit: 30000KTotal Submissions: 1820Accepted: 844DescriptionGiven a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement of these integers so that each adjacent pair of integers sums to a composite (non-prime) number 阅读全文

摘要： Prime LandTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 1630Accepted: 745DescriptionEverybody in the Prime Land is using a prime base number system. In this system, each positive integer x is represented as follows: Let {pi}i=0,1,2,... denote the increasing sequence of all prime numbers. 阅读全文

摘要： 最近在学习数论：贴一下Miller_Rabbin测试；代码：#include<stdio.h>#include<stdlib.h>#include<time.h>#define maxTest 100__int64 Random(__int64 n){ return (__int64)((double)rand()/RAND_MAX*n+0.5);}__int64 Modular_Exp(__int64 a, __int64 b, __int64 n) // a^b mod n { __int64 ans; if(b == 0) return 1; if(b 阅读全文

摘要： 直播入口——欢迎围观ACM/ICPC北京现场赛~ Computer TransformationTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2921Accepted Submission(s): 1111Problem DescriptionA sequence consisting of one digit, the number 1 is initially written into a computer. At each succes 阅读全文

摘要： 直播入口——欢迎围观ACM/ICPC北京现场赛~ Ignatius and the Princess IITime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1873Accepted Submission(s): 1129Problem DescriptionNow our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is a 阅读全文