hdu1852 Beijing 2008 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others) Total Submission(s): 502 Accepted Submission(s): 172 Problem Description As we all know, the next Olympic Games will be held in Beijing in 2008. So…

A sequence of numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4550    Accepted Submission(s): 1444 Problem Description Xinlv wrote some sequences on the paper a long time ago, they might…

题目链接 Problem Description Galen Marek, codenamed Starkiller, was a male Human apprentice of the Sith Lord Darth Vader. A powerful Force-user who lived during the era of the Galactic Empire, Marek originated from the Wookiee home planet of Kashyyyk as…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1576 Problem Description 要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973) = 1). Input 数据的第一行是一个T,表示有T组数据.每组数据有两个数n(0 <= n < 9973)和B(1 <= B <= 10^9). Output 对应每组数据输出(A/B)%9973. Sample…

Sum Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=4704 Mean: 给定一个大整数N,求1到N中每个数的因式分解个数的总和. analyse: N可达10^100000,只能用数学方法来做. 首先想到的是找规律.通过枚举小数据来找规律,发现其实answer=pow(2,n-1); 分析到这问题就简单了.由于n非常大,所以这里要用到费马小定理:a^n ≡ a^(n%(m-1)) * a^(m-1)≡ a^(n%(m-…

Tom and matrix Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=5226 Mean: 题意很简单,略. analyse: 直接可以用Lucas定理+快速幂水过的,但是我却作死的用了另一种方法. 方法一:Lucas定理+快速幂水过 方法二:首先问题可以转化为求(0,0),(n,m)这个子矩阵的所有数之和.画个图容易得到一个做法,对于n<=m,答案就是2^0+2^1+...+2^m=2^(m+1)-1,对于n>m…

http://acm.hdu.edu.cn/showproblem.php?pid=6395 Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1475    Accepted Submission(s): 539 Problem Description Let us define a sequence as belo…

题目链接 题意: 思路: 直接拿别人的图,自己写太麻烦了~ 然后就可以用矩阵快速幂套模板求递推式啦~ 另外: 这题想不到或者不会矩阵快速幂,根本没法做,还是2013年长沙邀请赛水题,也是2008年Google Codejam Round 1A的C题. #include <bits/stdc++.h> typedef long long ll; const int N = 5; int a, b, n, mod; /* *矩阵快速幂处理线性递推关系f(n)=a1f(n-1)+a2f(n-2)+.…

http://poj.org/problem?id=1995 题意:求(A1^B1 + A2^B2 + .....Ah^Bh)%M 直接快速幂,以前对快速幂了解不深刻,今天重新学了一遍so easy 以a^b为例:如果b是偶数那么一定可以写成 (a^2 * a^2 ....)一共是b/2个,那么其实就可以写成(a*a)^(b/2),另a = a*a,b= b/2,此时还是求a^b,只不过a和b已经变了,但是没有问题,还是可以按照上面的方法在判断的,如果b是奇数的话就把a的一个给拿出来先与ans相…

Rightmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 57430    Accepted Submission(s): 21736 Problem Description Given a positive integer N, you should output the most right digit of N…