题意: 定义 \[f(n)=\sum\limits_{i=1}^{n-1}(i\oplus (n-i))\] 求\(f(n),n \leq 10^{500}\) 分析: 这个数列对应OEIS的A006582 先上公式: \[f(n)=\left\{\begin{matrix} 4f(k)+6k,n=2k+1\\ 2f(k)+2f(k-1)+4k-4,n=2k \end{matrix}\right.\] 递推的思路就是虽然不知道两个数的异或值,但是如果知道这两个数的奇偶性那么结果的奇偶性也就知道了…

Exclusive or 题目链接: http://acm.hust.edu.cn/vjudge/contest/121336#problem/J Description Given n, find the value of Note: ♁ denotes bitwise exclusive-or. Input The input consists of several tests. For each tests: A single integer n (2≤n<10^500). Output…

题意很简单, 就是给个n, 算下面这个式子的值. $\sum\limits_{i=1}^{n-1} i \otimes (n-i)$ 重点是n的范围:2≤n<10500 比赛的时候 OEIS一下得到了一个公式: $a_0=a_1=a_2=0$; n为偶数 : $2 \times a_{\frac{n}{2}}+2 \times a_{\frac{n}{2}-1}+4\times (\frac{n}{2}-1) $ n为奇数 : $4\times a_{\frac{n-1}{2}}+6\times…

Exclusive or Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 327    Accepted Submission(s): 137 Problem Description Given n, find the value of  Note: ⊕ denotes bitwise exclusive-or.   Input T…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4816 Problem Description The Bathysphere is a spherical deep-sea submersible which was unpowered and lowered into the ocean on a cable, and was used to conduct a series of dives under the sea. The Bathys…

LCM Walk Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5584 Description A frog has just learned some number theory, and can't wait to show his ability to his girlfriend. Now the frog is sitting on a grid map o…

题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=4336 题意简单,直接用容斥原理即可 AC代码: #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cstdlib> #include <cmath> #include <vector> #include &…

balls Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5570 Description There are n balls with m colors. The possibility of that the color of the i-th ball is color j is ai,jai,1+ai,2+...+ai,m. If the number of b…

意甲冠军:那是,  从数0-n小球进入相应的i%a箱号.然后买一个新的盒子. 今天的总合伙人b一个盒子,Bob试图把球i%b箱号. 求复位的最小成本. 每次移动的花费为y - x ,即移动前后盒子编号的差值的绝对值. 算法: 题目就是要求                  watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvdTAxMjg0MTg0NQ==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissol…

链接:pid=4790">http://acm.hdu.edu.cn/showproblem.php?pid=4790 意:从[a.b]中随机找出一个数字x,从[c.d]中随机找出一个数字y.给出p.m,假设(x+y)%p==m则算成功,问成功的概率是多少. 思路:[a.b]中连续p个数.[c,d]中连续p个数.用这2*p个数进行组合能找到p种的成功组合(详细不证),所以找到[a.b]中p循环的个数x1,[c,d]中p循环的个数y1,则它们组成的成功组合数为p*x1*y1. 然后是处理边界…