简单的题.直接找题意来就好了. #include <iostream> #include <cstdio> using namespace std; int s, n, ans; int main() { cin >> s >> n; n -= s; * ) ans++; * ) ans++; * ) ans++; ) ans++; cout<<ans<<endl; }…

SGU 127 题意:给你n个数字,和m,k,问你有多少个数字的m次幂可以被k整除 收获:快速幂 #include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(…

About this Course This course will teach you how to build convolutional neural networks and apply it to image data. Thanks to deep learning, computer vision is working far better than just two years ago, and this is enabling numerous exciting applica…

matlab基础教程--根据Andrew Ng的machine learning整理 基本运算 算数运算 逻辑运算 格式化输出 小数位全局修改 向量和矩阵运算 矩阵操作 申明一个矩阵或向量 快速建立一个矩阵或向量 随机矩阵方阵生成 magic矩阵生成(每行每列相加和相同) 获取矩阵的维度size 获取矩阵的最大维度length 矩阵操作.获取单个元素.行.列.赋值 矩阵append.矩阵元素放到一个列向量中 矩阵运算 矩阵乘法 A*C:根据矩阵乘法公式相乘. A .* B:矩阵元素对应相乘. 矩…

题目大意: 给出平面上若干个点的坐标,你的任务是建一个环形围墙,把所有的点围在里面,且距所有点的距离不小于l.求围墙的最小长度. 思路: 很容易得出答案就是凸包周长+以l为半径的圆的周长. 这里讲一下Andrew算法. Andrew是Graham算法的变种,而且Andrew更快,更稳定. Andrew算法思想是先将n个点按照x坐标从小到大排序(x相同按照y从小到大),得到一个序列a1,a2,...an,将a1,a2放入ch数组,从a3开始,判断点是否在凸包当前前进方向的左边,如果是,就将点加入c…

Hanoi Tower Troubles Again! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 602    Accepted Submission(s): 418 Problem Description People stopped moving discs from peg to peg after they know the…

链接:ZOJ1239 Hanoi Tower Troubles Again! Description People stopped moving discs from peg to peg after they know the number of steps needed to complete the entire task. But on the other hand, they didn't not stopped thinking about similar puzzles with…

水概率....SGU里难得的水题.... 495. Kids and Prizes Time limit per test: 0.5 second(s)Memory limit: 262144 kilobytes input: standardoutput: standard ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the…

sgu 101 - Domino Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u Description Dominoes – game played with small, rectangular blocks of wood or other material, each identified by a number of dots, or pips, on its face. The bl…

http://acm.sgu.ru/problem.php?contest=0&problem=495 题意:N个箱子M个人,初始N个箱子都有一个礼物,M个人依次等概率取一个箱子,如果有礼物则拿出礼物放回盒子,如果没有礼物则不操作.问M个人拿出礼物个数的期望.(N,M<=100000) #include <cstdio> using namespace std; double mpow(double a, int n) { double r=1; while(n) { if(n&…