HDU5407 CRB and Candies 题意: 计算\(LCM(C(n,0),C(n,1),C(n,2),\cdots,C(n,n-1),C(n,n))\) \(n\le 10^6\) 题解: 规律: \(LCM(C(n,0),C(n,1),C(n,2),\cdots,C(n,n-1),C(n,n)) = \frac{LCM(1,2,3,\cdots,,n-1,n,n+1)}{n+1}\) 令\(g(n)=LCM(C(n,0),C(n,1),C(n,2),\cdots,C(n,n-1),…

Problem Description CRB has N different candies. He is going to eat K candies.He wonders how many combinations he can select.Can you answer his question for all K (0 ≤ K ≤ N )?CRB is too hungry to check all of your answers one by one, so he only asks…

官方题解: The problem is just to calculate g(N) = LCM(C(N,0),C(N,1),...,C(N,N)) Introducing function f(n) = LCM(1,2,...,n), the fact g(n) = f(n+1)/(n+1) holds. We calculate f(n) in the following way. f(1)=1 If n =p^k,then f(n) = f(n−1)× p, else f(n) = f(…

pid=5407">[HDOJ 5407] CRB and Candies 赛后看这题题解仅仅有满眼的迷茫------ g(N) = LCM(C(N,0),C(N,1),...,C(N,N)) f(n)\ =\ LCM(1, 2, ..., n)f(n) = LCM(1,2,...,n), the fact g(n)\ =\ f(n+1) / (n+1)g(n) = f(n+1)/(n+1) f(n)\ =\ LCM(1, 2, ..., n)f(1) = 1 If n\ =p^{k}n …

CRB and Candies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 722    Accepted Submission(s): 361 Problem Description CRB has N different candies. He is going to eat K candies.He wonders how ma…

题目 CRB and Candies 题意 \[ \text{给定正整数N,求} LCM \lbrace C \left(N , 0 \right),C\left(N , 1 \right),...,C\left(N , N \right) \rbrace \% mod \qquad 1\leq N \leq 10^6 \] 题解 根据规律推出公式,另外关于这个公式的证明 \[ LCM \lbrace C \left(N , 0 \right),C\left(N , 1 \right),...,…

题目链接: Hdu 5407 CRB and Candies 题目描述: 给出一个数n,求lcm(C(n,0),C[n,1],C[n-2]......C[n][n-2],C[n][n-1],C[n][n])%(1e9+7)是多少? 解题思路: 刚开始的时候各种开脑洞,然后卡题卡的风生水起.最后就上了数列查询这个神奇的网站,竟然被我找到了!!!!就是把题目上给的问题转化为求lcm(1, 2, 3, 4 ...... n-2, n-1, n, n-1) / (n+1),扎扎就打了两个表一个lcm[n…

CRB and Candies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 453    Accepted Submission(s): 222 Problem Description   CRB has N different candies. He is going to eat K candies.He wonders how…

题目传送门 题意:求LCM (C(N,0),C(N,1),...,C(N,N)),LCM是最小公倍数的意思,C函数是组合数. 分析:先上出题人的解题报告 好吧,数论一点都不懂,只明白f (n + 1)意思是前n+1个数的最小公倍数,求法解释参考HDOJ 1019,2028 这个结论暂时不知道怎么推出来的,那么就是剩下1/(n+1) 逆元的求法了 代码: /************************************************ * Author :Running_Time…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5407 题目: Problem Description CRB has N different candies. He is going to eat K candies.He wonders how many combinations he can select.Can you answer his question for all K(0 ≤ K ≤ N)?CRB is too hungry to…