http://poj.openjudge.cn/practice/C18H

题目

算平均数用到公式\[\bar{x}=\frac{x_1+x_2+x_3+\cdots+x_n}{n}\]

但如果用int型计算，那么\(x_1+x_2+x_3+\cdots+x_n\)可能会超过\(2^{31}-1\)

算6个数的平均数可以这么算

Calculate the average of\(x_1,x_2,x_3\)
\[\bar{x}_1=\frac{x_1+x_2+x_3}{3}\]
Calculate the average of\(x_4,x_5,x_6\)
\[\bar{x}_2=\frac{x_4+x_5+x_6}{3}\]
Calculate the average of\(\bar{x}_1,\bar{x}_2\)
\[\bar{x}=\frac{\bar{x}_1+\bar{x}_2}{2}\]
In this way, as you can see, we actually add up at most $3$ integers at one time, instead of adding all the $6$ integers together. Therefore, as long as all the integers are not greater than \(\left\lfloor {\left( {{2^{31}} - 1} \right)/3} \right\rfloor \), we are not at risk of getting an overflow result. Thus, we call the value $71582782$ the Safe Upper Bound of $6$.

输入N，输出N的安全上界

题解

某日无聊翻openjudge的poj队伍，发现了PKU的校赛，想找一道最简单的题满足虚荣心:(

看了好久没看懂在干什么，看样例用计算器猜是$2^{31}-1$除以N的最大素因子

数论不行:(

照着书抄了个Pollard Rho+Miller-Rabin算法 TLE（其实根本就不知道复杂度

于是尝试Eratosthenes线性筛……可是需要开的数组太大……貌似无解了

其实还是自己太菜:(

为什么可以这么做呢……猜可能和这个过程有关
\[\left\lfloor {\frac{{a + b}}{2}} \right\rfloor  = \frac{{a + b}}{2} - \frac{{\left( {a + b} \right)\% 2}}{2}\]
\[\left\lfloor {\frac{{c + d}}{2}} \right\rfloor  = \frac{{c + d}}{2} - \frac{{\left( {c + d} \right)\% 2}}{2}\]
\[\left\lfloor {\frac{{a + b}}{2}} \right\rfloor  + \left\lfloor {\frac{{c + d}}{2}} \right\rfloor  = \frac{{a + b + c + d}}{2} - \frac{{a\% 2 + b\% 2 + c\% 2 + d\% 2}}{2}\]
\[\left\lfloor {\frac{{\left\lfloor {\frac{{a + b}}{2}} \right\rfloor  + \left\lfloor {\frac{{c + d}}{2}} \right\rfloor }}{2}} \right\rfloor  = \frac{{a + b + c + d}}{4} - \frac{{a\% 2 + b\% 2 + c\% 2 + d\% 2}}{4}\]
至于\(\frac{{a\% 2 + b\% 2 + c\% 2 + d\% 2}}{4}\)是否等于\({\left( {a + b + c + d} \right)\% 4}\)

我还是菜鸟，等以后变强了再看看……推广也只有以后了

空间问题抄了UESTC大神的代码

https://vjudge.net/solution/15934751

看了以后感觉自己真的太菜了:(

这差距不是一点啊……还得加油

原文地址：https://www.cnblogs.com/sahdsg/p/10369063.html