要算的数组命名为data var sum = function(x,y){ return x+y;}; //求和函数 var square = function(x){ return x*x;}; //数组中每个元素求它的平方 var data = [1,1,3,5,5]; // var mean = data.reduce(sum)/data.length; var deviations = data.map(function(x){return x-mean;}); var stddev =
//求n个数中的最小k个数 public static void TestMin(int k, int n) { Random rd = new Random(); int[] myArray = new int[n]; int[] newArray = new int[k]; for (int i = 0; i < n; i++) { // rand
[本文出自天外归云的博客园] 题1:求m以内的素数(m>2) def find_all_primes_in(m): def prime(num): for i in range(2, num): if divmod(num, i)[1] == 0: return False return True print([i for i in range(2, m + 1) if prime(i)]) if __name__ == '__main__': find_all_primes_in(100) 我
//求两个数中不同的位的个数 #include <stdio.h> int count_different(int a, int b) { int count = 0; int c = a^b; //a,b中不同的位即为1 while (c) { count++; c = c&(c - 1); //把c中最后一个1去掉 } return count; } int main() { printf("%d\n", count_different(3,8)); //3 p
package wac.wev.as;//新建一个方法在求最大值import java.util.Scanner; public class MaxLian {public static void main(String[] args){//键盘录入以及导包Scanner sc= new Scanner(System.in);//数据接收System.out.println("请输入第一个数据:");int a = sc.nextInt();System.out.println(&qu
pog loves szh II Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2106 Accepted Submission(s): 606 Problem Description Pog and Szh are playing games.There is a sequence with n numbers, Pog wi
275. To xor or not to xor The sequence of non-negative integers A1, A2, ..., AN is given. You are to find some subsequence Ai 1, Ai 2, ..., Ai k (1 <= i 1 < i 2 < ... < i k<= N) such, that Ai 1 XOR Ai 2 XOR ... XOR Ai k has a maximum valu
Console.WriteLine("请输入第一个数:"); int a = Convert.ToInt32( Console.ReadLine()); Console.WriteLine("请输入第二个数:"); int b = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("请输入第三个数:"); int c = Convert.ToInt32(Console.ReadLine(
Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 5646 Accepted: 1226 Description In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p: ⊕ is the xor operator. We say a path the xor-l
#include<stdio.h> int max(int a,int b,int c); int main() { int a,b,c; while(scanf("%d %d %d",&a,&b,&c)!=EOF){ printf("%d\n",max(a,b,c));} ; } int max(int a,int b,int c) { int m; m=a; if(b>m) m=b; if(c>m) m=c; re
17.4 Write a method which finds the maximum of two numbers. You should not use if-else or any other comparison operator. 这道题让我们找出两个数中的较大值,不能用if..else..语句判断,也不能用任何比较符号.那么我们怎么办呢,我们看两个数的差值a-b是否大于0,如果大于0,说明a大,如果小于0,说明b大.然后我们用一个变量k来记录a-b的符号位,用q来表示k的相反数,这样
