B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following proble…

Codeforces Round #262 (Div. 2) B B - Little Dima and Equation B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot a…

B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following proble…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment. Find all…

http://codeforces.com/contest/460/problem/B import java.util.*; import java.math.*; public class Main { public static void main(String []args) { Scanner cin=new Scanner(System.in); int a,b,c; BigInteger a1="); BigInteger a10="); BigInteger a3=&q…

题目链接:http://codeforces.com/problemset/problem/460/B 题目意思:给出a, b, c三个数,要你找出所有在 1 ≤ x ≤ 1e9 范围内满足 x = b·s(x)a + c  这条等式的x的个数,并输出相应的 x 具体是多少. 不看tutorial 都不知道,原来枚举的方向错了,人家是枚举1-81 的情况,我就是枚举1-1e9, = =...直接暴力即可,有个比较要注意的地方,算方程右边的时候有可能超过int,需要用long long 或 __i…

题目地址:http://codeforces.com/contest/460/problem/B 这题乍一看没思路.可是细致分析下会发现,s(x)是一个从1到81的数,不管x是多少.所以能够枚举1到81,这样就转化成了一个一元一次方程.直接求解x就能够了.这时候还要推断x是否在1到10^9之间,而且它的各位数之和是s(x). 这题脑残了两次.. . 第一次是写成了<=10^9..然后发现错误后,就又改了回来. .可是发现10^9是不可能的,. ... 代码例如以下: #include <ios…

Codeforces#262_1002 B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him…

题目链接 A. Vasya and Socks time limit per test:2 secondsmemory limit per test:256 megabytesinput:standard inputoutput:standard output Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. Wh…

题目链接 B Little Dima and Equation 题意:给a, b,c 给一个公式,s(x)为x的各个位上的数字和,求有多少个x. 分析:直接枚举x肯定超时,会发现s(x)范围只有只有1-81,所以枚举一下就行. 在做题的时候,用了pow()错了3次,反正以后不用pow了,还是手写吧.会有误差.pow返回的是double型的. 昨天在b题耽误了好多时间,先是提交错第一组,然后又被人cha了.注意在x在1-10^9之间. #include <iostream> #include &…

题目: B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following pr…

Little Dima and Equation Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 460B Description Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the…

详见:http://robotcator.logdown.com/posts/221514-codeforces-round-262-div-2 1:A. Vasya and Socks   http://codeforces.com/contest/460/problem/A 有n双袜子,每天穿一双然后扔掉.每隔m天买一双新袜子,问最多少天后没有袜子穿. . 简单思维题:曾经不注重这方面的训练,结果做了比較久.这样的题自己边模拟边想.只是要多考虑trick ```c++ int main(){…

A. Vasya and Socks time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When…

 FZU 2102   Solve equation Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice Description You are given two positive integers A and B in Base C. For the equation: A=k*B+d We know there always existing many non-nega…

题意: 有1~9数字各有a1, a2, -, a9个, 有无穷多的+和=. 问只用这些数字, 最多能组成多少个不同的等式x+y=z, 其中x,y,z∈[1,9]. 等式中只要有一个数字不一样 就是不一样的 思路: 计算下可以发现, 等式最多只有36个. 然后每个数字i的上界是17-i个 可以预先判掉答案一定是36的, 然后直接暴力搜索每个等式要不要就好了. 注意剪枝即可 ; int a[maxn]; bool flag36; int ans; struct Equation { int x, y…

#对coursera上Andrew Ng老师开的机器学习课程的笔记和心得: #注:此笔记是我自己认为本节课里比较重要.难理解或容易忘记的内容并做了些补充,并非是课堂详细笔记和要点: #标记为<补充>的是我自己加的内容而非课堂内容,参考文献列于文末.博主能力有限,若有错误,恳请指正: #---------------------------------------------------------------------------------# 多元线性回归的模型: #-----------…

,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, 1 % Exercise 1: Linear regression with multiple variables %% Initialization %% ================ Part 1: Featu…

Can you solve this equation? Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Now,given the equation *x^ + *x^ + *x^ + *x + == Y,can you find its solution between and ; No…

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27938#problem/E 题目大意:Given, n, count the number of solutions to the equation x+2y+2z=n, where x,y,z,n are non negative integers. 解题思路:只对一个枚举由此推算出另外两个的种类,千万不要都枚举!!! #include<iostream> #include<…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7156    Accepted Submission(s): 3318 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…

题目链接: http://codeforces.com/problemset/problem/272/D D. Dima and Two Sequences time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, …

H - The equation Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u Submit Status Practice SGU 106 Description There is an equation ax + by + c = 0. Given a,b,c,x1,x2,y1,y2 you must determine, how many integer roots of this eq…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7493    Accepted Submission(s): 3484 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,…

继续考虑Liner Regression的问题,把它写成如下的矩阵形式,然后即可得到θ的Normal Equation. Normal Equation: θ=(XTX)-1XTy 当X可逆时,(XTX)-1XTy = X-1,(XTX)-1XTy其实就是X的伪逆(Pseudo inverse).这也对应着Xθ = y ,θ = X-1y 考虑特殊情况 XTX 不可逆 解决办法: 1)考虑是否有冗余的特征,例如特征中有平方米,还有平方厘米,这两个特征就是冗余的,解决办法是去掉冗余 2)再有就是n…

http://acm.hdu.edu.cn/howproblem.php?pid=2199 Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13468    Accepted Submission(s): 6006 Problem Description Now,given the…

蓝桥杯--Quadratic Equation 问题描述 求解方程ax2+bx+c=0的根.要求a, b, c由用户输入,并且可以为任意实数. 输入格式:输入只有一行,包括三个系数,之间用空格格开. 输出格式:输出只有一行,包括两个根,大根在前,小根在后,无需考虑特殊情况,保留小数点后两位. 输入输出样例样例输入2.5 7.5 1.0样例输出-0.14 -2.86 java code: import java.util.*;import java.text.*;public class Yiyu…

Can you solve this equation? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6023    Accepted Submission(s): 2846 Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y…

D. Dima and Lisa Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/584/problem/D Description Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them…

The Solutions of Nonlinear Equation 本文主要介绍几种用于解非线性方程$f(x)=0$的一些方法. (1) Bisection Method. 算法: step 1: 初始化$a,b(b>a)$,使$f(a),f(b)$异号. step 2: while (停止条件不满足) $p=a+\frac{b-a}{2}$: 若 $f(p)f(a)<0$,$b=p$:否则$a=p$. end while step 3: 返回的$p$为方程$f(x)=0$的解. 停止条件…