Work out the first ten digits of the sum of the following one-hundred 50-digit numbers. 371072875339021027987979982208375902465101357402504637693767749000971264812489697007805041701826053874324986199524741059474233309513058123726617309629919422133635…

聪明的办法是想:求前10位,那只要前8位加起来,进2位就OK. 本的办法,就是真的加起来,截前面10位.如我. numList = str.split() sum = 0 for i in range(0,len(numList)): sum += int(numList[i][0:50]) print i,numList[i][0:50],sum print sum Work out the first ten digits of the sum of the following one-hu…

题意:计算出以下一百个50位数的和的前十位数字. /************************************************************************* > File Name: euler013.c > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > Created Time: 2017年06月25日 星期日 10时55分56秒 *******************…

Text Aeroplanes are slowly driving me mad. I live near an airport and passing planes can be heard night and day. The airport was built years ago, but for some reason it could not be used then. Last year, however, it came into use. Over a hundred peop…

原题链接:http://ac.jobdu.com/problem.php?pid=1554 由数列的前缀和:$\begin{align*}\Large{} S_n &=\Large{}\sum_{i=1}^{n}{{a_i}} \ \ \ \ i=1,2,3...n\end{align*}$ 由于:$\begin{align*}\Large{} S_n -S_{n-1}&=\Large{}a_n \end{align*}$ 所以区间$\begin{align*}\Large{} [i,j]…

Triple Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 388    Accepted Submission(s): 148 Problem Description Given the finite multi-set A of n pairs of integers, an another finite multi-set B …

论搜索顺序的优越性. 从小到大搜索,T到怀疑人生. 从大到小搜索,轻松AC. 双向搜索. 是这样的.我们首先排序一下,然后翻转.(当然也可以写cmp) 然后我们搜前半部分,把结果记录在某一数组中. 然后把结果排序(去重),再搜另一部分. 接下来有两种想法: 1.同样方式处理另一部分,然后双指针扫描. 2.第二次每次搜到终点,二分更新ans. 还可以加上剪枝:如果当前的和加上后面所有再加上第一部分最大值都达不到ans,剪枝. 还可以用set,但是CE了...不知道为什么. 细节顺序还可以更优,但是…

If you’re like most homeowners, purchasing a home represents the single biggest financial transaction of your lifetime. And two of the most important payments leading up to the closing of your new home are the deposit and down payment. As a first-tim…

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=4008 一道不简单的概率和期望dp题 根据期望的线性性质,容易想到,可以算出每张卡的期望伤害,然后全部加在一起 手算样例之后发现是正确的,那么我们只要求出每张卡的实际被使用的概率就可以了 记第$i$张卡的实际被使用的概率为$fp[i]$ 那么答案就是 $\Large\sum\limits_{i=0}^{n-1}fp[i]\cdot d[i]$ 如何求出$fp[i]$? 首先考虑第一张卡的$f…

https://eoscity.io/f/viewtopic.php?f=7&t=17 这篇文章的原文:   (https://steemit.com/eos/@genereos/eos-multisig-tutorial) What is Multisig Multisig is a term frequently used in the cryptographic and blockchain space. Essentially it enables multiple parties to…