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…

题目链接:http://codeforces.com/problemset/problem/509/D 题意:题目给出公式w[i][j]= (a[i] + b[j])% k; 给出w,要求是否存在这样的数列,若存在则求出a,b 和k 题解:如果有符合条件的点的话那么a[i],b[j]可以任意转换比如说a[i],b[j]可以转化为a[i]-p,b[i]+p.所以只要存在 符合条件的解的a[i]可以为任意值.也就是说a[i]可以先赋值为0然后其他就都可以推出来了,当然开始推出后会发现 有负的,没事,…

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…

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…

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] = …

[题目]B. Recover the String [题意]找到一个串s,满足其中子序列{0,0}{0,1}{1,0}{1,1}的数量分别满足给定的数a1~a4,或判断不存在.数字<=10^9,答案<=10^6. [算法]数学构造 [题解]首先由a1和a4易得0的数量x0和1的数量x1. 容易发现01和10关系密切,令1的位置为b1...bx1,则: {0,1} (b1-1)+(b2-2)+(b3-3)+...+(bx1-x1)=a2 {1,0} (x0-b1+1)+(x0-b2+1)+...…

http://codeforces.com/problemset/problem/716/C codeforces716C. Plus and Square Root 这个题就是推,会推出来规律,发现和等级有关系,等级的平方然后减去数字啥的,忘记了,推的草稿纸找不到了... 推出来就是类似a+b的题目. 代码: 1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath&…

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…

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…