题目链接 题意 要找两个合数,使他们两个的差为\(n\),\(n\)为题目给出的数 思路 我们可以枚举减数\(now\),判断一下是不是质数,如果是质数就让\(now++\),然后用一个数\(tot\)记录被减数,也就是\(now\)加\(n\),判断\(tot\)是不是质数,如果是质数再让\(now++\),如果不是质数我们就找到了答案,因为我们已经保证了\(now\)为合数,所以就可以直接跳出循环输出答案啦~ 因为数据范围并不大,所以我们可以直接判断一个数是不是质数,如下 //判断是否为质数…

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…

 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: #-----------…

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…

,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, 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…