Sicily1020-大数求余算法及优化】的更多相关文章

Github最终优化代码: https://github.com/laiy/Datastructure-Algorithm/blob/master/sicily/1020.c 题目如下: 1020. Big Integer Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description Long long ago, there was a super computer that could deal with VeryLongInt…
题意:给出一个大数,这个大数由两个素数相乘得到,让我们判断是否其中一个素数比L要小,如果两个都小,输出较小的那个. 分析:大数求余的方法:针对题目中的样例,143 11,我们可以这样算,1 % 11 = 1:      1×10 + 4 % 11 = 3:      3×10 + 3 % 11 = 0;我们可以把大数拆成小数去计算,同余膜定理保证了这个算法的这正确性,而且我们将进制进行一定的扩大也是正确的. 注意:素数打标需要优化,否则超时.   进制需要适当,100和1000都可以,10进制超…
题意: 项的自幂级数求和为 11 + 22 + 33 + - + 1010 = 10405071317. 求如下一千项的自幂级数求和的最后10位数字:11 + 22 + 33 + - + 10001000. 思路: 求最后十位数字 % 1010 即可. 对于快速幂中数据溢出的问题,有两种解决方法: 1. 方法一:对于两个数 x y,现在想求 x * y % MOD,可以将 x 表示成 a * DIGS + b,y 表示成 c * DIGS + d,x * y % MOD 则等价与 ( a * c…
题目链接:http://poj.org/problem?id=2635 题目分析: http://blog.csdn.net/lyy289065406/article/details/6648530…
Large Division Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c. Input Input starts with an integer T…
Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c. Input Input starts with an integer T (≤ 525), denot…
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Problem Descripton Two planets named Haha and Xixi in the universe and they were created with the universe beginning. There is 73 days in Xixi a year and 137 days in Haha a year. Now you know the days N after Big Bang, you need to answer whether it i…
Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5930    Accepted Submission(s): 4146 Problem Description As we know, Big Number is always troublesome. But it's really important in our…
做TopCoder SRM 576 D2 L3 题目时,程序有个地方需要对一个数大量求幂并取余,导致程序运行时间很长,看了Editoral之后,发现一个超级高效的求幂并取余的算法,之前做System test时,程序运行时间(最慢的测试用例)为500ms左右,使用此方法之后,运行时间直接减为20ms,快了20多倍,所以将此方法记录下来. 算法时间复杂度为 log(n). 这个算法其实就是  数据结构与算法分析 (Weiss 著) 一书中开头的那个递归求幂算法的非递归版,简洁明了. 代码如下: /…