题目链接 Problem Description Amtel has announced that it will release a 128-bit computer chip by 2010, a 256-bit computer by 2020, and so on, continuing its strategy of doubling the word-size every ten years. (Amtel released a 64-bit computer in 2000, a…

http://poj.org/problem?id=2661 题意:Amtel在1960年发行了4位计算机,并实行每十年位数翻一番的策略,将最大整数n作为改变的等级,其中n!表示计算机的无符号整数(n!<=无符号整数).给出一个年份1960<=y<=2160,求此时计算机的等级n是多少? 思路:首先求出计算机在第y年的位数k,则此时计算机的无符号整数为2^k-1,则 n!<=2^k-1, 直接求n的阶乘容易溢出,此时可以两边同时取对数得: log2(n!)<=log2(2^k…

Factstone Benchmark Amtel has announced that it will release a 128-bit computer chip by 2010, a 256-bit computer by 2020, and so on, continuing its strategy of doubling the word-size every ten years. (Amtel released a 64-bit computer in 2000, a 32-bi…

[问题描述] Amtel已经宣布,到2010年,它将发行128位计算机芯片:到2020年,它将发行256位计算机:等等,Amtel坚持每持续十年将其字大小翻一番的战略.(Amtel于2000年发行了64位计算机,在1990年发行了32位计算机,在1980年发行了16位计算机,在1970年发行了8位计算机,并首先在1960年发行了4位计算机) Amtel将使用新的标准检查等级——Factstone——来宣传其新芯片大大提高的能力. Factstone等级被定义为这样的最大整数n,使得n!可以表示为…

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…