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A. Plus and Square Root time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ZS the Coder is playing a game. There is a number displayed on the screen and there are two buttons, ' + ' (plus) an…
Square root digital expansion It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all. The square root of two i…
E. Square Root of Permutation A permutation of length n is an array containing each integer from 1 to n exactly once. For example, q = [4, 5, 1, 2, 3] is a permutation. For the permutation q the square of permutation is the permutation p that p[i] = …
C. Plus and Square Root time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ZS the Coder is playing a game. There is a number displayed on the screen and there are two buttons, ' + ' (plus) an…
It is possible to show that the square root of two can be expressed as an infinite continued fraction.  2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213... By expanding this for the first four iterations, we get: 1 + 1/2 = 3/2 = 1.51 + 1/(2 + 1/2) = 7…
Square RootWhen the square root functional configuration is selected, a simplified CORDIC algorithm isused to calculate the positive square root of the input. The input, X_IN, and the output,X_OUT, are always positive and are both expressed as either…
A. Plus and Square Root 题目连接: http://codeforces.com/contest/715/problem/A Description ZS the Coder is playing a game. There is a number displayed on the screen and there are two buttons, ' + ' (plus) and '' (square root). Initially, the number 2 is d…
ZS the Coder is playing a game. There is a number displayed on the screen and there are two buttons, ' + ' (plus) and '' (square root). Initially, the number 2 is displayed on the screen. There are n + 1 levels in the game and ZS the Coder start at t…
Tiny Wong the chef used to be a mathematics teacher in a senior high school. At that time, he always used to tell his students that when there is a square root of some number in one’s final result, it should be simplified by factoring out the largest…
五十七.平方根收敛(Square root convergents) 二的平方根可以表示为以下这个无穷连分数: \[ \sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}} \] 通过把前四项展开,我们得到: \[ \begin{aligned}1 + \frac 1 2 &= \frac 32 = 1.5\\1 + \frac 1 {2 + \frac 1 2} &= \frac 7 5 = 1.4\\1 + \frac 1 {2…
A permutation of length n is an array containing each integer from 1 to n exactly once. For example, q = [4, 5, 1, 2, 3] is a permutation. For the permutation q the square of permutation is the permutation p that p[i] = q[q[i]] for each i = 1... n. F…
不理解,背板子 #include<cstdio> using namespace std; int Pow(int a,int b,int p) { ; ) ) res=1LL*a*res%p; return res; } bool Legendre(int a,int p) { >>,p)==; } void modsqr(int a,int p) { int x; int i,k,b; ) x=a%p; ==) x=Pow(a,p+>>,p); else { ;Le…
http://acm.timus.ru/problem.aspx?space=1&num=1132 题意: 求 x^2 ≡ n mod p  p是质数 的 解 本题中n>=1 特判p=2,接下来求当p是奇素数时的解 引理1: 引理2:方程有解当且仅当 定理: 设a满足 不是模p的二次剩余, 即无解, 那么是二次剩余方程的解 #include<cstdio> #include<cstdlib> #include<algorithm> using namesp…
传送门 MD写一道二次剩余的板题差点写自闭了. 我用的是cipollacipollacipolla算法. 利用的是欧拉准则来找寻一个二次非剩余类来求根. 注意这题有两个等根和模数为2的情况. 代码: #include<bits/stdc++.h> using namespace std; typedef long long ll; int T,n,mod; inline int ksm(int a,int p){int ret=1;for(;p;p>>=1,a=(ll)a*a%mo…
Description A permutation of length n is an array containing each integer from 1 to n exactly once. For example, q = [4, 5, 1, 2, 3] is a permutation. For the permutation q the square of permutation is the permutation p that p[i] = q[q[i]] for each i…
A permutation of length n is an array containing each integer from 1 to n exactly once. For example, q = [4, 5, 1, 2, 3] is a permutation. For the permutation q the square of permutation is the permutation p that p[i] = q[q[i]] for each i = 1... n. F…
题意:求${x^2} \equiv n\bmod p$ 解题关键: 定理:若$a$满足$w = {a^2} - n$是模$p$的二次非剩余,即,${x^2} = w\bmod p$无解,则${(a + \sqrt w )^{\frac{{p + 1}}{2}}}$是二次剩余方程${x^2} \equiv n\bmod p$的解. 证明: $\begin{array}{l}{x^2} \equiv {(a + \sqrt w )^{p + 1}} \equiv (a + \sqrt w ){(a…
题意: 求\(x^2 \equiv a \mod p\) 的所有整数解 思路: 二次剩余定理求解. 参考: 二次剩余Cipolla's algorithm学习笔记 板子: //二次剩余,p是奇质数 ll ppow(ll a, ll b, ll mod){ ll ret = 1; a = a % mod; while(b){ if(b & 1) ret = ret * a % mod; a = a * a % mod; b >>= 1; } return ret; } struct TT…
题目链接 分析:这题都过了2000了,应该很简单..写这篇只是为了凑篇数= = 假设在第级的时候开方过后的数为,是第级的系数.那么 - 显然,最小的情况应该就是, 化简一下公式,在的情况下应该是,注意时特殊情况,应该为2. /*****************************************************/ //#pragma comment(linker, "/STACK:1024000000,1024000000") #include <map>…
题目链接 Java程序 package projecteuler51to60; import java.math.BigInteger; import java.util.Iterator; import java.util.Set; import java.util.TreeSet; class level57{ void solve0(){ /*** a a+2b --- ------- b a+b ***/ int count = 0; BigInteger a=BigInteger.va…
from decimal import getcontext, Decimal def main(): n = int(raw_input()) p = int(raw_input()) getcontext().prec = p+10 # 扩大精度,保证接过 sum = 0 for i in range(1,n+1): nTemp = Decimal(i).sqrt() if nTemp._isinteger() : # 自生函数的判断 sum += 0 else: nTemp = str(n…
牛顿法(Newton’s method)又称为牛顿-拉弗森法(Newton-Raphson method),是一种近似求解实数方程式的方法.(注:Joseph Raphson在1690年出版的<一般方程分析>中提出了后来被称为“牛顿-拉弗森法”的数学方法,牛顿于1671年写成的著作<流数法>中亦包括了这个方法,但该书在1736年才出版.) 之前的一篇博客中提到的二分法可以求解方根(用二分法定义平方根函数),而使用牛顿迭代法可以更快地解出方根.现在,人们使用的计算器里面大多数都是运用…
Python里面有内置(Built-in)的平方根函数:sqrt(),可以方便计算正数的平方根.那么,如果要自己定义一个sqrt函数,该怎么解决呢? 解决思路:  1. 大于等于1的正数n的方根,范围肯定在0~n之间:小于1的正数n的方根,范围肯定在0~1之间  2. 用二分法(Bisection method, Binary search)从中间开始找n的方根.  3. 对于大于等于1的正数n,先假设n/2是n的方根,如果n/2的平方大于n,那么说明n的方根在0~n/2之间:如果n/2的平方小…
题目链接 \(Description\) 给定一个\(n\)的排列\(p_i\),求一个排列\(q_i\),使得对于任意\(1\leq i\leq n\),\(q_{q_i}=p_i\).无解输出\(-1\). \(1\leq n\leq10^6\). \(Solution\) 对排列\(q_i\)我们建一张图,边为\(i\to q_i\).显然这张图是由几个环构成. 发现对于\(q_{q_i}\)的图,原来\(q_i\)中的奇环它们还是类似的一个奇环,原来的偶环会分裂成两个大小相等的偶环. 所…
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public class Solution { public static void main(String[] args) { Scanner ip = new Scanner(System.in); System.out.print("Enter a number: "); double n = ip.nextDouble(); System.out.println(sqrt(n)); ip.close(); } public static double sqrt(double n…
题目分析 我们首先模拟一下题意 假设有一个 \(q _1\) \(p\) \(a_1\) \(a_x\) \(a_{a_1}\) \(a_{a_x}\) \(q\) \(x\) \(a_1\) \(a_x\) \(a_{a_1}\) \(pos\) 1 \(x\) \(a_1\) \(a_x\) 对这个表格分析,发现,当前节点的后继为前驱的对应 如果我们对 \(i \to p_i\) 建边,模拟一下,发现 偶环是会有两个环组成,而奇环会形成一个类似于五角星的图形 于是,我们可以找出所有环,对于奇…
Input The input stream contains a set of integer numbers Ai (0 ≤ Ai ≤ 1018). The numbers are separated by any number of spaces and line breaks. A size of the input stream does not exceed 256 KB. Output For each number Ai from the last one till the fi…
The problem is so easy, that the authors were lazy to write a statement for it! Input The input stream contains a set of integer numbers Ai (0 ≤ Ai ≤ 10^18). The numbers are separated by any number of spaces and line breaks. A size of the input strea…
The problem is so easy, that the authors were lazy to write a statement for it! Input The input stream contains a set of integer numbers Ai (0 ≤ Ai ≤ 1018). The numbers are separated by any number of spaces and line breaks. A size of the input stream…